NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

<div class=”entry-content” style=”height: auto !important;”>Ex 9.1 Class 11 Maths 
Sequences and Series Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series All Exercises were prepared by Experienced LearnCBSE.in Teachers.
Free download NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1, Ex 9.2, Ex 9.3, Ex 9.4 and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.
<ul>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-9-sequences-and-series/#Sequences_and_Series_Class_11_Ex_9.1″>Sequences and Series Class 11 Ex 9.1</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-9-sequences-and-series/#Sequences_and_Series_Class_11_Ex_9.2″>Sequences and Series Class 11 Ex 9.2</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-9-sequences-and-series/#Sequences_and_Series_Class_11_Ex_9.3″>Sequences and Series Class 11 Ex 9.3</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11-maths-chapter-9-sequences-and-series-ex-9-4/#Sequences_and_Series_Class_11_Ex_9.4″>Sequences and Series Class 11 Ex 9.4</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-9-sequences-and-series/#Sequences_and_Series_Class_11_Miscellaneous_Exercise”>Sequences and Series Class 11 Miscellaneous Exercise</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-9-sequences-and-series/#अनुक्रम_तथा_श्रेणी_प्रश्नावली_9.1_का_हल_हिंदी_में”>अनुक्रम तथा श्रेणी प्रश्नावली 9.1 का हल हिंदी में</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11-maths-chapter-9-ex-9-2/#अनुक्रम_तथा_श्रेणी_प्रश्नावली_9.2_का_हल_हिंदी_में”>अनुक्रम तथा श्रेणी प्रश्नावली 9.2 का हल हिंदी में</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11-maths-chapter-9-ex-9-3/#अनुक्रम_तथा_श्रेणी_प्रश्नावली_9.3_का_हल_हिंदी_में”>अनुक्रम तथा श्रेणी प्रश्नावली 9.3 का हल हिंदी में</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11-maths-chapter-9-sequences-and-series-ex-9-4/#अनुक्रम_तथा_श्रेणी_प्रश्नावली_9.4_का_हल_हिंदी_में”>अनुक्रम तथा श्रेणी प्रश्नावली 9.4 का हल हिंदी में</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11-maths-chapter-9-miscellaneous-exercise/#अनुक्रम_तथा_श्रेणी_विविध_प्रश्नावली_का_हल_हिंदी_में”>अनुक्रम तथा श्रेणी विविध प्रश्नावली का हल हिंदी में</a></li>
<li><a href=”https://www.learncbse.in/sequences-and-series-class-11-notes/”>Sequences and Series Class 11 Notes</a></li>
<li><a href=”https://www.learncbse.in/ncert-exemplar-problems-class-11-mathematics-chapter-9-sequence-series/”>NCERT Exemplar Class 11 Maths Sequences and Series</a></li>
<li><a href=”https://www.learncbse.in/jee-main-previous-year-papers-questions-with-solutions-maths-sequences-and-series/”>JEE Main Mathematics Sequences and Series Previous Year Questions</a></li>
</ul>
<h2>NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series</h2>
<img loading=”lazy” class=”size-full wp-image-90176 aligncenter” src=”https://www.learncbse.in/wp-content/uploads/2019/08/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series” width=”672″ height=”378″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/08/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series.png 672w, https://www.learncbse.in/wp-content/uploads/2019/08/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-300×169.png 300w” sizes=”(max-width: 672px) 100vw, 672px”>
Topics and Sub Topics in Class 11 Maths Chapter 9 Sequences and Series:
<table border=”2″>
<tbody>
<tr>
<td width=”126″>Section Name</td>
<td width=”438″>Topic Name</td>
</tr>
<tr>
<td width=”126″>9</td>
<td width=”438″>Sequences and Series</td>
</tr>
<tr>
<td width=”126″>9.1</td>
<td width=”438″>Introduction</td>
</tr>
<tr>
<td width=”126″>9.2</td>
<td width=”438″>Sequences</td>
</tr>
<tr>
<td width=”126″>9.3</td>
<td width=”438″>Series</td>
</tr>
<tr>
<td width=”126″>9.4</td>
<td width=”438″>Arithmetic Progression (A.P.)</td>
</tr>
<tr>
<td width=”126″>9.5</td>
<td width=”438″>Geometric Progression (G.P.)</td>
</tr>
<tr>
<td width=”126″>9.6</td>
<td width=”438″>Relationship Between A.M. and G.M.</td>
</tr>
<tr>
<td width=”126″>9.7</td>
<td width=”438″>Sum to n terms of Special Series</td>
</tr>
</tbody>
</table>
<h3 style=”text-align: center;”> NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1</h3>
Ex 9.1 Class 11 Maths Question 1: 
Write the first five terms of the sequence whose nth term is an = n (n + 2). 
Ans: 
an = n(n +2) 
Substituting n = 1, 2, 3, 4 and 5, we obtain 
a1 = 1 (1 + 2) = 3, 
a2 = 2 (2 + 2) = 8, 
a3 = 3 (3 + 2) = 15, 
a4 = 4 (4 + 2) = 24, 
a5 = 5 (5 + 2) = 35 
Therefore, the required terms are 3, 8, 15, 24 and 35.
Ex 9.1 Class 11 Maths Question 2: 
Write the first five terms of the sequence whose nth term is an = nn+1<script type=”math/tex” id=”MathJax-Element-1″>\frac{n}{n+1}</script>. 
Ans:<div class=”google-auto-placed ap_container” style=”width: 100%; height: auto; clear: both; text-align: center;”><ins data-ad-format=”auto” class=”adsbygoogle adsbygoogle-noablate” data-ad-client=”ca-pub-7398766921532682″ data-adsbygoogle-status=”done” style=”display: block; margin: auto; background-color: transparent; height: 280px;” data-ad-status=”filled”><div id=”aswift_2_host” style=”border: none; height: 280px; width: 750px; margin: 0px; padding: 0px; position: relative; visibility: visible; background-color: transparent; display: inline-block; overflow: visible;” tabindex=”0″ title=”Advertisement” aria-label=”Advertisement”><iframe id=”aswift_2″ name=”aswift_2″ style=”left:0;position:absolute;top:0;border:0;width:750px;height:280px;” sandbox=”allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation” width=”750″ height=”280″ frameborder=”0″ marginwidth=”0″ marginheight=”0″ vspace=”0″ hspace=”0″ allowtransparency=”true” scrolling=”no” src=”https://googleads.g.doubleclick.net/pagead/ads?client=ca-pub-7398766921532682&output=html&h=280&adk=1731550593&adf=3055222944&pi=t.aa~a.1381849204~i.20~rp.4&w=750&fwrn=4&fwrnh=100&lmt=1671089791&num_ads=1&rafmt=1&armr=3&sem=mc&pwprc=3683862639&ad_type=text_image&format=750×280&url=https%3A%2F%2Fwww.learncbse.in%2Fncert-solutions-for-class-11th-maths-chapter-9-sequences-and-series%2F&fwr=0&pra=3&rh=188&rw=750&rpe=1&resp_fmts=3&wgl=1&fa=27&adsid=ChAIgJK1nQYQsIu6srmxoPNtEj0Ai1qrVRMfpaIZYfQpIlSvZ1OoWxiheTWOo715d630l9Zt5QX24urtNdt_WxNnNZWJSQtYw12haH60Ta8z&uach=WyJXaW5kb3dzIiwiMTAuMC4wIiwieDg2IiwiIiwiMTA4LjAuNTM1OS4xMjUiLFtdLGZhbHNlLG51bGwsIjY0IixbWyJOb3Q_QV9CcmFuZCIsIjguMC4wLjAiXSxbIkNocm9taXVtIiwiMTA4LjAuNTM1OS4xMjUiXSxbIkdvb2dsZSBDaHJvbWUiLCIxMDguMC41MzU5LjEyNSJdXSxmYWxzZV0.&dt=1672336380031&bpp=8&bdt=3077&idt=8&shv=r20221207&mjsv=m202212070101&ptt=9&saldr=aa&abxe=1&cookie=ID%3D70e399fdacbd4ad1-22148803a5d8002c%3AT%3D1669478827%3ART%3D1669478827%3AS%3DALNI_MYm9yd6L8vzb5BDmbgSwdsFdiI7aQ&gpic=UID%3D00000b83c07f2775%3AT%3D1669478827%3ART%3D1672321466%3AS%3DALNI_Mb7gJ5SG0-pM1C2eCtov6mlwxKKVQ&prev_fmts=0x0%2C728x280%2C1005x124&nras=3&correlator=7358255305457&frm=20&pv=1&ga_vid=1355004618.1669478872&ga_sid=1672336379&ga_hid=694764345&ga_fc=1&u_tz=330&u_his=1&u_h=768&u_w=1366&u_ah=728&u_aw=1366&u_cd=24&u_sd=1&dmc=4&adx=105&ady=2672&biw=1349&bih=657&scr_x=0&scr_y=51&eid=44759875%2C44759926%2C44759842%2C44773809%2C31071251%2C44777949%2C44779794%2C44780792&oid=2&psts=ACgb8tsfqPevzYJyw-cJjlgBgAa2_0ikq1-ul90ykSKt4GwpCpQcJmE01U0SDxJ24E-nKYhiQNbyQ-rCyqPclJFy%2CACgb8ttZ3-iC7smczW8jYqFwmePOlqbbqdIlIyv-hK7NsyB0w8funWL5qdX79LUQuu8XH6Dfct48g8ry7-sNf0X56wD4bibqVXnpEvbps0LxVaw&pvsid=2544932135427130&tmod=777830844&uas=0&nvt=1&eae=0&fc=1408&brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C657&vis=1&rsz=%7C%7Cs%7C&abl=NS&fu=128&bc=31&ifi=3&uci=a!3&btvi=2&fsb=1&xpc=An7cqWghUN&p=https%3A//www.learncbse.in&dtd=M” data-google-container-id=”a!3″ data-google-query-id=”CIeXrvGyn_wCFQK4aAodgk8JzQ” data-load-complete=”true”></iframe></div></ins></div>
an = nn+1<script type=”math/tex” id=”MathJax-Element-2″>\frac{n}{n+1}</script> 
Sustituting n = 1, 2, 3, 4, 5, we otain 
an = 11+1=12<script type=”math/tex” id=”MathJax-Element-3″>\frac{1}{1+1}=\frac{1}{2}</script>
an = 22+1=23<script type=”math/tex” id=”MathJax-Element-4″>\frac{2}{2+1}=\frac{2}{3}</script>
an = 33+1=34<script type=”math/tex” id=”MathJax-Element-5″>\frac{3}{3+1}=\frac{3}{4}</script>
an = 44+1=45<script type=”math/tex” id=”MathJax-Element-6″>\frac{4}{4+1}=\frac{4}{5}</script>
an = 55+1=56<script type=”math/tex” id=”MathJax-Element-7″>\frac{5}{5+1}=\frac{5}{6}</script>
Therefore, the required terms are 12<script type=”math/tex” id=”MathJax-Element-8″>\frac{1}{2}</script>, 23<script type=”math/tex” id=”MathJax-Element-9″>\frac{2}{3}</script> , 34<script type=”math/tex” id=”MathJax-Element-10″>\frac{3}{4}</script> , 45<script type=”math/tex” id=”MathJax-Element-11″>\frac{4}{5}</script> and 56<script type=”math/tex” id=”MathJax-Element-12″>\frac{5}{6}</script> .
More Resources for CBSE Class 11
<ul>
<li><a href=”https://www.learncbse.in/rd-sharma-class-11-solutions-chapter-21-special-series/”>RD Sharma Class 11 Solutions Some Special Series</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-2/”>NCERT Solutions</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11-maths/”>NCERT Solutions Class 11 Maths</a></li>
<li><a href=”https://www.learncbse.in/15381-2/ncert-solutions-class-11-physics/”>NCERT Solutions Class 11 Physics </a></li>
<li><a href=”https://www.learncbse.in/15381-2/ncert-solutions-class-11-chemistry/”>NCERT Solutions Class 11 Chemistry</a></li>
<li><a href=”https://www.learncbse.in/15381-2/ncert-solutions-for-class-11-biology/”>NCERT Solutions Class 11 Biology</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-class-11-hindi-core/”>NCERT Solutions Class 11 Hindi</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11-english/”>NCERT Solutions Class 11 English</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-for-class-11-business-studies/”>NCERT Solutions Class 11 Business Studies</a></li>
<li><a href=”https://www.learncbse.in/ncert-solutions-class-11-computer-science-python/”>NCERT Solutions Class 11 Computer Science</a></li>
</ul>
Ex 9.1 Class 11 Maths Question 3: 
Write the first five terms of the sequence whose nth term is an = 2n. 
Ans: 
an = 2n 
Substituting n = 1, 2, 3, 4, 5, we obtain 
a1 = 21 = 2 
a2 = 22 = 4 
a3 = 23 = 8 
a4 = 24 = 16 
a5 = 25 = 32 
Therefore, the required terms are 2, 4, 8, 16 and 32.
Ex 9.1 Class 11 Maths Question 4: 
Write the first five terms of the sequence whose nth term is an = 2n−36<script type=”math/tex” id=”MathJax-Element-13″>\frac{2 n-3}{6}</script>. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120391″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q4.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q4.1″ width=”432″ height=”324″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q4.1.png 432w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q4.1-300×225.png 300w” sizes=”(max-width: 432px) 100vw, 432px”>
Ex 9.1 Class 11 Maths Question 5: 
<img loading=”lazy” class=”alignnone size-full wp-image-120393″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q5.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q5″ width=”494″ height=”35″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q5.png 494w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q5-300×21.png 300w” sizes=”(max-width: 494px) 100vw, 494px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120392″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q5.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q5.1″ width=”509″ height=”235″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q5.1.png 509w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q5.1-300×139.png 300w” sizes=”(max-width: 509px) 100vw, 509px”>
Ex 9.1 Class 11 Maths Question 6: 
Write the first five terms of the sequence whose nth term is an = nn2+54<script type=”math/tex” id=”MathJax-Element-14″>n \frac{n^{2}+5}{4}</script>. 
Ans: 
Substituting n = 1, 2, 3, 4, 5, we obtain 
<img loading=”lazy” class=”alignnone size-full wp-image-138817″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 1″ width=”475″ height=”211″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-1.png 475w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-1-300×133.png 300w” sizes=”(max-width: 475px) 100vw, 475px”>
Ex 9.1 Class 11 Maths Question 7: 
Find the indicated terms in the following sequence whose nth term is an =4n – 3; a17, a24. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120394″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q7.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q7.1″ width=”247″ height=”152″>
Ex 9.1 Class 11 Maths Question 8: 
<img loading=”lazy” class=”alignnone size-full wp-image-120396″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q8.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q8″ width=”478″ height=”51″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q8.png 478w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q8-300×32.png 300w” sizes=”(max-width: 478px) 100vw, 478px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120395″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q8.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q8.1″ width=”236″ height=”92″>
Ex 9.1 Class 11 Maths Question 9: 
Find the indicated term in the following sequence whose nt1 term is an = (- 1)n – 1 n3; a9. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120397″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q9.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q9.1″ width=”227″ height=”82″>
Ex 9.1 Class 11 Maths Question 10: 
<img loading=”lazy” class=”alignnone size-full wp-image-120399″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q10.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q10″ width=”526″ height=”57″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q10.png 526w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q10-300×33.png 300w” sizes=”(max-width: 526px) 100vw, 526px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120398″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q10.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q10.1″ width=”238″ height=”92″>
Ex 9.1 Class 11 Maths Question 11: 
Write the first five terms of the following sequence and obtain the corresponding series: 
a1 = 3, an = 3an – 1 + 2 for all n > 1. 
Ans:
a1 = 3, an = 3 an – 1 + 2 for all n > 1 
=> a2 = 3a2 – 1 + 2 
= 3a1 + 2 
= 3(3) + 2 = 11
a3 = 3a3 – 1 + 2 
= 3a2 + 2 
= 3(11) + 2 = 35
a4 = 3a4 – 1 + 2 
= 3a3 + 2 
= 3(35) + 2 = 107
a5 = 3a5 – 1 + 2 
= 3a4 + 2 
= 3(107) + 2 = 323. 
Hence, the first five terms of the sequence are 3, 11, 35, 107 and 323. 
The corresponding series is 3 + 11 + 35 + 107 + 323 + …
Ex 9.1 Class 11 Maths Question 12: 
<img loading=”lazy” class=”alignnone size-full wp-image-120401″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q12.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q12″ width=”576″ height=”79″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q12.png 576w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q12-300×41.png 300w” sizes=”(max-width: 576px) 100vw, 576px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120400″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q12.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Q12.1″ width=”545″ height=”356″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q12.1.png 545w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-Q12.1-300×196.png 300w” sizes=”(max-width: 545px) 100vw, 545px”>
Ex 9.1 Class 11 Maths Question 13: 
Write the first five terms of the following sequence and obtain the corresponding series: 
a1 = a2 = 2, an = an – 1 – 1, n > 2. 
Ans:
a1 = a2 = 2, 
an = an – 1, n > 2 
=> a3 = a2 – 1 = 2 – 1 = 1, 
a4 = a3 – 1 = 1 – 1 = 0 
a5 = a4 – 1 = 0 – 1 = – 1. 
Hence, the first five terms of the sequence are 2, 2, 1, 0 and – 1. 
The corresponding series is 2 + 2 +1 + 0+ (- 1) + ……….<div class=”google-auto-placed ap_container” style=”width: 100%; height: auto; clear: both; text-align: center;”><ins data-ad-format=”auto” class=”adsbygoogle adsbygoogle-noablate” data-ad-client=”ca-pub-7398766921532682″ data-adsbygoogle-status=”done” style=”display: block; margin: auto; background-color: transparent; height: 280px;”><div id=”aswift_3_host” style=”border: none; height: 280px; width: 750px; margin: 0px; padding: 0px; position: relative; visibility: visible; background-color: transparent; display: inline-block;”></div></ins></div>
Ex 9.1 Class 11 Maths Question 14: 
The Fibonacci sequence is defined by 1 = a1 = a2 and an = an – 1 + an – 2 n > 2. Find an+1an<script type=”math/tex” id=”MathJax-Element-15″>\frac{\boldsymbol{a}_{n+1}}{\boldsymbol{a}_{n}}</script>, for n = 1, 2, 3, 4, 5. 
Ans:
1 = a1 = a2 
an = an – 1 + an – 2, n > 2 
a3 = a2 + a1 
= 1 + 1 = 2, 
a4 = a3 + a2 
= 2 + 1 = 3, 
a5 = a4 + a3 
= 3 + 2 = 5, 
a6 = a5 + a3 
= 5 + 3 = 8<div class=”google-auto-placed ap_container” style=”width: 100%; height: auto; clear: both; text-align: center;”><ins data-ad-format=”auto” class=”adsbygoogle adsbygoogle-noablate” data-ad-client=”ca-pub-7398766921532682″ data-adsbygoogle-status=”done” style=”display: block; margin: auto; background-color: transparent; height: 280px;”><div id=”aswift_4_host” style=”border: none; height: 280px; width: 750px; margin: 0px; padding: 0px; position: relative; visibility: visible; background-color: transparent; display: inline-block;”></div></ins></div>
For n = 1, 
an+1an=a2a1=11<script type=”math/tex” id=”MathJax-Element-16″>\frac{a_{n+1}}{a_{n}}=\frac{a_{2}}{a_{1}}=\frac{1}{1}</script> = 1
For n = 2, 
an+1an=a3a2=21<script type=”math/tex” id=”MathJax-Element-17″>\frac{a_{n+1}}{a_{n}}=\frac{a_{3}}{a_{2}}=\frac{2}{1}</script> = 2
For n = 3, 
an+1an=a4a3=32<script type=”math/tex” id=”MathJax-Element-18″>\frac{a_{n+1}}{a_{n}}=\frac{a_{4}}{a_{3}}=\frac{3}{2}</script>
For n = 4, 
an+1an=a5a4=53<script type=”math/tex” id=”MathJax-Element-19″>\frac{a_{n+1}}{a_{n}}=\frac{a_{5}}{a_{4}}=\frac{5}{3}</script>
For n = 5, 
an+1an=a6a5=85<script type=”math/tex” id=”MathJax-Element-20″>\frac{a_{n+1}}{a_{n}}=\frac{a_{6}}{a_{5}}=\frac{8}{5}</script>
<h3><a id=”अनुक्रम_तथा_श्रेणी_प्रश्नावली_9.1_का_हल_हिंदी_में”></a>NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (अनुक्रम तथा श्रेणी) Hindi Medium Ex 9.1</h3>
<img loading=”lazy” class=”alignnone size-large wp-image-91643″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-1-791×1024.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.1″ width=”791″ height=”1024″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-1-791×1024.png 791w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-1-232×300.png 232w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-1-768×994.png 768w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-1.png 1700w” sizes=”(max-width: 791px) 100vw, 791px”> 
<img loading=”lazy” class=”alignnone size-full wp-image-91644″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-2.png” alt=”11 Maths Chapter 9 Sequences and Series Exercise 9.1 in Hindi Medium” width=”532″ height=”692″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-2.png 532w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-2-231×300.png 231w” sizes=”(max-width: 532px) 100vw, 532px”> 
<img loading=”lazy” class=”alignnone size-full wp-image-91645″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-3.png” alt=”11 Maths Chapter 9 Exercise 9.1 in Hindi Medium” width=”532″ height=”642″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-3.png 532w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-3-249×300.png 249w” sizes=”(max-width: 532px) 100vw, 532px”> 
<img loading=”lazy” class=”alignnone size-full wp-image-91646″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-4.png” alt=”Class 11 Maths Chapter 9 Optional Exercise 9.1 all question answers in hindi” width=”529″ height=”551″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-4.png 529w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-4-288×300.png 288w” sizes=”(max-width: 529px) 100vw, 529px”> 
<img loading=”lazy” class=”alignnone wp-image-91647 size-full” src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-5.png” alt=”Class 11 Maths Chapter 9 Optional Exercise 9.1 for up, gujrat, mp board cbse” width=”525″ height=”677″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-5.png 525w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-5-233×300.png 233w” sizes=”(max-width: 525px) 100vw, 525px”> 
<img loading=”lazy” class=”alignnone size-full wp-image-91648″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-6.png” alt=”Class 11 Maths Chapter 9 Optional Exercise 9.1 download free pdf” width=”531″ height=”383″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-6.png 531w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.1-6-300×216.png 300w” sizes=”(max-width: 531px) 100vw, 531px”>
<h3 style=”text-align: center;”>NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2</h3>
Ex 9.2 Class 11 Maths Question 1: 
Find the sum of odd integers from 1 to 2001. 
Ans: 
The odd integers from 1 to 2001 are 1, 3, 5, … 1999, 2001. 
This sequence forms an A.P. 
Here, first term, a = 1 
Common difference, d = 2 
Here, a + (n – 1) d = 2001 
⇒ 1 + (n -1) (2) = 2001 
⇒ 2n – 2 = 2000 
⇒ n = 1001 
Sn = n2<script type=”math/tex” id=”MathJax-Element-21″>\frac{n}{2}</script> [2a + (n – 1)d] 
= 10012<script type=”math/tex” id=”MathJax-Element-22″>\frac{1001}{2}</script> [2 × 1 + (1001 – 1) × 2] 
= 10012<script type=”math/tex” id=”MathJax-Element-23″>\frac{1001}{2}</script> [2 + 1000 x 2] 
= 10012<script type=”math/tex” id=”MathJax-Element-24″>\frac{1001}{2}</script> × 2002 
= 1001 × 1001 
= 1002001 
Thus, the sum of odd numbers from 1 to 2001 is 1002001.
Ex 9.2 Class 11 Maths Question 2: 
<img loading=”lazy” class=”alignnone size-full wp-image-120403″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q2″ width=”587″ height=”43″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q2.png 587w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q2-300×22.png 300w” sizes=”(max-width: 587px) 100vw, 587px”> 
Ans: 
The natural numbers lying between 100 and 1000, which are multiples of 5 are 105, 110, …, 995. 
Here, a = 105 and d = 5 
a + (n – 1) d = 995 
⇒ 105 + (n -1)5 = 995 
⇒ (n – 1)5 = 995 – 105 = 890 
⇒ n – 1 = 178 
⇒ n = 179 
Sn = 1792<script type=”math/tex” id=”MathJax-Element-25″>\frac{179}{2}</script> [2 (105) + (179 – 1) (5)] 
= 1792<script type=”math/tex” id=”MathJax-Element-26″>\frac{179}{2}</script> [2 (105) + (178) (5)] 
= 179 [105 + (89)5] 
= (179) (105 + 445) 
= (179) (550) = 98450 
Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5 is 98450.<div class=”google-auto-placed ap_container” style=”width: 100%; height: auto; clear: both; text-align: center;”><ins data-ad-format=”auto” class=”adsbygoogle adsbygoogle-noablate” data-ad-client=”ca-pub-7398766921532682″ data-adsbygoogle-status=”done” style=”display: block; margin: auto; background-color: transparent; height: 280px;”><div id=”aswift_5_host” style=”border: none; height: 280px; width: 750px; margin: 0px; padding: 0px; position: relative; visibility: visible; background-color: transparent; display: inline-block;”></div></ins></div>
Ex 9.2 Class 11 Maths Question 3: 
In an A.P. the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is – 112. 
Ans: 
Let the first term of given AP be a and common difference be d. 
We have, T1 = a = 2 
T1 + T2 + T3 + T4 + T5 = [T6 + T7 + T8 + T9 + T10] 
Sum of 5 terms, where first term is a = 14<script type=”math/tex” id=”MathJax-Element-27″>\frac{1}{4}</script> × sum of 5 terms, where first term is (a + 5d) 
⇒ 52<script type=”math/tex” id=”MathJax-Element-28″>\frac{5}{2}</script> [2a + (5 – 1) d] = 14<script type=”math/tex” id=”MathJax-Element-29″>\frac{1}{4}</script> × 52<script type=”math/tex” id=”MathJax-Element-30″>\frac{5}{2}</script> [2(a + 5d) + (5 – 1)d] 
[∵ Sn = n2<script type=”math/tex” id=”MathJax-Element-31″>\frac{n}{2}</script> [2a + (n – 1)d]] 
52<script type=”math/tex” id=”MathJax-Element-32″>\frac{5}{2}</script> [2a + 4d] = 14<script type=”math/tex” id=”MathJax-Element-33″>\frac{1}{4}</script> × 52<script type=”math/tex” id=”MathJax-Element-34″>\frac{5}{2}</script> [2a + 10d + 4d] 
52<script type=”math/tex” id=”MathJax-Element-35″>\frac{5}{2}</script> [2a + 4d] = 14<script type=”math/tex” id=”MathJax-Element-36″>\frac{1}{4}</script> × 52<script type=”math/tex” id=”MathJax-Element-37″>\frac{5}{2}</script> [2a + 14d] 
⇒ 2a + 4d = 14<script type=”math/tex” id=”MathJax-Element-38″>\frac{1}{4}</script> [2a + 14d] 
2(2) + 4d = 14<script type=”math/tex” id=”MathJax-Element-39″>\frac{1}{4}</script> [2 . (2) + 14d] [put a = 2] 
4 + 4d = 14<script type=”math/tex” id=”MathJax-Element-40″>\frac{1}{4}</script> [4 + 14d] 
16 + 16d = 4 + 14d 
16d – 14d = 4 – 16 
2d = – 12 
d = – 6 
T20 = a + (20 – 1) d 
= 2 + 19 × (- 6) 
= 2 – 114 = – 112 
Hence proved.<div class=”google-auto-placed ap_container” style=”width: 100%; height: auto; clear: both; text-align: center;”><ins data-ad-format=”auto” class=”adsbygoogle adsbygoogle-noablate” data-ad-client=”ca-pub-7398766921532682″ data-adsbygoogle-status=”done” style=”display: block; margin: auto; background-color: transparent; height: 280px;”><div id=”aswift_6_host” style=”border: none; height: 280px; width: 750px; margin: 0px; padding: 0px; position: relative; visibility: visible; background-color: transparent; display: inline-block;”></div></ins></div>
Ex 9.2 Class 11 Maths Question 4: 
How many terms of the A.P. – 6, – 112<script type=”math/tex” id=”MathJax-Element-41″>\frac{11}{2}</script>, – 5, … are needed to give the sum – 25? 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120404″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q4.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q4″ width=”559″ height=”522″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q4.png 559w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q4-300×280.png 300w” sizes=”(max-width: 559px) 100vw, 559px”>
Ex 9.2 Class 11 Maths Question 5: 
In an A.P., if pth term is 1q<script type=”math/tex” id=”MathJax-Element-42″>\frac{1}{q}</script> and qth term is 1p<script type=”math/tex” id=”MathJax-Element-43″>\frac{1}{p}</script>, prove that the sum of first pq terms is 12<script type=”math/tex” id=”MathJax-Element-44″>\frac{1}{2}</script> (pq + 1), where pq ≠ q. 
Ans: 
It is known that the general term of an A.P. is 
an = a + (n – 1) d 
According to the given information, 
pth term = ap 
= a + (p – 1)d = 1q<script type=”math/tex” id=”MathJax-Element-45″>\frac{1}{q}</script> …………..(i) 
qth term = aq 
= a + (q – 1)d = 1p<script type=”math/tex” id=”MathJax-Element-46″>\frac{1}{p}</script> …………….(ii) 
Subtracting eq. (ii) from eq. (i), we obtain 
(p – 1) d – (q – 1) d = 1q−1p<script type=”math/tex” id=”MathJax-Element-47″>\frac{1}{q}-\frac{1}{p}</script> 
(p – 1 – q + 1) d = p−qpq<script type=”math/tex” id=”MathJax-Element-48″>\frac{p-q}{p q}</script> 
⇒ (p – q) d = p−qpq<script type=”math/tex” id=”MathJax-Element-49″>\frac{p-q}{p q}</script> 
d = 1pq<script type=”math/tex” id=”MathJax-Element-50″>\frac{1}{p q}</script> 
Putting the value of d in eq. (i), we obtain 
<img loading=”lazy” class=”alignnone size-full wp-image-138822″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 1″ width=”357″ height=”238″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-1.png 357w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-1-300×200.png 300w” sizes=”(max-width: 357px) 100vw, 357px”><div class=”google-auto-placed ap_container” style=”width: 100%; height: auto; clear: both; text-align: center;”><ins data-ad-format=”auto” class=”adsbygoogle adsbygoogle-noablate” data-ad-client=”ca-pub-7398766921532682″ data-adsbygoogle-status=”done” style=”display: block; margin: auto; background-color: transparent; height: 280px;”><div id=”aswift_7_host” style=”border: none; height: 280px; width: 750px; margin: 0px; padding: 0px; position: relative; visibility: visible; background-color: transparent; display: inline-block;”></div></ins></div>
Ex 9.2 Class 11 Maths Question 6: 
If the sum of a certain number, of terms of the A.P. 25, 22, 19, …is 116. Find the last term. 
Ans: 
Let the sum of n terms of the given A.P. be 116. 
Sn = n2<script type=”math/tex” id=”MathJax-Element-51″>\frac{n}{2}</script> [2a + (n – 1)d] 
Here, a = 25 and d = 22 – 25 = – 3 
Sn = n2<script type=”math/tex” id=”MathJax-Element-52″>\frac{n}{2}</script> [2 × 25 + (n – 1) (- 3)] 
⇒ 116 = n2<script type=”math/tex” id=”MathJax-Element-53″>\frac{n}{2}</script> [50 – 3n + 3] 
⇒ 232 = n(53 – 3n) = 53n – 3n2 
3n2 – 24n – 29n + 232 = 0 
3n (n – 8) – 29 (n – 8) = 0 
(n – 8) (3n – 29) = 0 
n = 8 or n = 293<script type=”math/tex” id=”MathJax-Element-54″>\frac{29}{3}</script> 
However, n cannot be equal to 293<script type=”math/tex” id=”MathJax-Element-55″>\frac{29}{3}</script>. 
Therefore, n = 8 
∴ a8 = Last term = a + (n -1) d = 25 + (8 – 1) (- 3) 
= 25 + (7) (- 3) = 25 – 21 = 4 
Thus, the last term of the A.P. is 4.
Ex 9.2 Class 11 Maths Question 7: 
<img loading=”lazy” class=”alignnone size-full wp-image-120405″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q7.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q7.1″ width=”400″ height=”34″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q7.1.png 400w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q7.1-300×26.png 300w” sizes=”(max-width: 400px) 100vw, 400px”> 
Ans: 
It is given that the kth term of the A.P. is 5k + 1 
kth term = ak = a + (k – 1) d 
a + (k – 1) d = 5k +1 
a + kd – d = 5k + 1 
Comparing the coefficient of k, we obtain d = 5 
a – d = 1 
⇒ a – 5 = 1 
⇒ a = 6 
Sn = n2<script type=”math/tex” id=”MathJax-Element-56″>\frac{n}{2}</script> [2a + (n – 1) d] 
= n2<script type=”math/tex” id=”MathJax-Element-57″>\frac{n}{2}</script> [2 (6) + (n – 1) (5)] 
= n2<script type=”math/tex” id=”MathJax-Element-58″>\frac{n}{2}</script> [12n + 5n – 5] 
= n2<script type=”math/tex” id=”MathJax-Element-59″>\frac{n}{2}</script> (5n + 7).
Ex 9.2 Class 11 Maths Question 8: 
<img loading=”lazy” class=”alignnone size-full wp-image-120408″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q8.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q8″ width=”543″ height=”49″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q8.png 543w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q8-300×27.png 300w” sizes=”(max-width: 543px) 100vw, 543px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120407″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q8.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q8.1″ width=”461″ height=”408″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q8.1.png 461w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q8.1-300×266.png 300w” sizes=”(max-width: 461px) 100vw, 461px”>
Ex 9.2 Class 11 Maths Question 9: 
The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120409″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q9.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q9.1″ width=”574″ height=”251″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q9.1.png 574w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q9.1-300×131.png 300w” sizes=”(max-width: 574px) 100vw, 574px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120410″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q9.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q9.2″ width=”402″ height=”303″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q9.2.png 402w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q9.2-300×226.png 300w” sizes=”(max-width: 402px) 100vw, 402px”>
Ex 9.2 Class 11 Maths Question 10: 
If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find sum of the first (p + q) terms. 
Ans: 
Let a and b be the first term and the common difference of the A.P. respectively. 
Here, Sp = p2<script type=”math/tex” id=”MathJax-Element-60″>\frac{p}{2}</script> [2a + (p – 1)d] 
Sq = q2<script type=”math/tex” id=”MathJax-Element-61″>\frac{q}{2}</script> [2a + (q – 1 )d] 
According to the given condition, 
p2<script type=”math/tex” id=”MathJax-Element-62″>\frac{p}{2}</script> [2a + (p – 1) d] = q2<script type=”math/tex” id=”MathJax-Element-63″>\frac{q}{2}</script> [2a + (q – 1) d] 
p [2a + (p – 1) d] = q [2a + (n – 1) d] 
2ap + pd (p – 1) d = 2aq + qd (q – 1) d 
2a (p – q) + d[p(p – 1) – q (q – 1)] = 0 
2a (p – q) + d[p2 – p – q2 + q] = 0 
2a (p – q) + d [(p – q) (p + q) – (p – q)] = 0 
2a (p – q) + d [(p – q) (p + q – 1)] = 0 
2a + d (p + q – 1) = 0 
d = −2ap+q−1<script type=”math/tex” id=”MathJax-Element-64″>\frac{-2 a}{p+q-1}</script> ……………..(i) 
∴ Sp + q = p+q2<script type=”math/tex” id=”MathJax-Element-65″>\frac{p+q}{2}</script> [2a + (p + q – 1) d] 
Sp + q = p+q2[2a+(p+q−1)(−2ap+q−1)]<script type=”math/tex” id=”MathJax-Element-66″>\frac{p+q}{2}\left[2 a+(p+q-1)\left(\frac{-2 a}{p+q-1}\right)\right]</script> [from eq. (i)] 
= p+q2<script type=”math/tex” id=”MathJax-Element-67″>\frac{p+q}{2}</script> [2a – 2a] = 0 
Thus, the sum of the first (p + q) terms of the A.P. is 0.
Ex 9.2 Class 11 Maths Question 11: 
<img loading=”lazy” class=”alignnone size-full wp-image-120413″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q11.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q11″ width=”463″ height=”92″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q11.png 463w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q11-300×60.png 300w” sizes=”(max-width: 463px) 100vw, 463px”> 
Ans: 
Given that, Sp = a, Sq = b, Sc = r 
Let A be the first term and d be the common difference. Then, 
Sp = p2<script type=”math/tex” id=”MathJax-Element-68″>\frac{p}{2}</script> [2A + (p – 1) d] = a 
2A + (p – 1) d = 2ap<script type=”math/tex” id=”MathJax-Element-69″>\frac{2 a}{p}</script> ……………(i) 
<img loading=”lazy” class=”alignnone size-full wp-image-138823″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 2″ width=”560″ height=”381″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-2.png 560w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-2-300×204.png 300w” sizes=”(max-width: 560px) 100vw, 560px”> <img loading=”lazy” class=”alignnone size-full wp-image-138824″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-3.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 3″ width=”555″ height=”204″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-3.png 555w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-3-300×110.png 300w” sizes=”(max-width: 555px) 100vw, 555px”><div class=”google-auto-placed ap_container” style=”width: 100%; height: auto; clear: both; text-align: center;”><ins data-ad-format=”auto” class=”adsbygoogle adsbygoogle-noablate” data-ad-client=”ca-pub-7398766921532682″ data-adsbygoogle-status=”done” style=”display: block; margin: auto; background-color: transparent; height: 280px;”><div id=”aswift_8_host” style=”border: none; height: 280px; width: 750px; margin: 0px; padding: 0px; position: relative; visibility: visible; background-color: transparent; display: inline-block;”></div></ins></div>
Ex 9.2 Class 11 Maths Question 12: 
<img loading=”lazy” class=”alignnone size-full wp-image-120414″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q12.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q12″ width=”526″ height=”45″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q12.png 526w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q12-300×26.png 300w” sizes=”(max-width: 526px) 100vw, 526px”> 
Ans: 
Let the first term “be a and common difference be d. 
Thus, Sm = m2<script type=”math/tex” id=”MathJax-Element-70″>\frac{m}{2}</script> [2a + (m – 1)d] and Sn = n2<script type=”math/tex” id=”MathJax-Element-71″>\frac{n}{2}</script> [2a + (n – 1)d] 
According to the given condition. 
SmSn=m2n2<script type=”math/tex” id=”MathJax-Element-72″>\frac{S_{m}}{S_{n}}=\frac{m^{2}}{n^{2}}</script> 
m2[2a+(m−1)d]n2[2a+(n−1)d]=m2n2⇒2a+(m−1)d2a+(n−1)d=mn<script type=”math/tex” id=”MathJax-Element-73″>\frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) d]}=\frac{m^{2}}{n^{2}} \Rightarrow \frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n}</script> 
2an + (mn – n) d = 2am + (mn – m)d 
2an – 2am = (mn – m – mn + n)d 
2a (n – m) = d (n – m) 
⇒ d = 2a . 
Tm = a + (m – 1) d = a + (m – 1) 2a 
Tm = a + 2am – 2a 
Tm = 2am – a 
⇒ Tm = a (2m – 1) ……………..(i) 
Also, Tn = a (2n – 1) ………………(ii) 
On dividing eQuestion (i) by eQuestion (ii), we get 
TmTn=a(2m−1)a(2n−1)=2m−12n−1<script type=”math/tex” id=”MathJax-Element-74″>\frac{T_{m}}{T_{n}}=\frac{a(2 m-1)}{a(2 n-1)}=\frac{2 m-1}{2 n-1}</script> 
Hence proved.
Ex 9.2 Class 11 Maths Question 13: 
<img loading=”lazy” class=”alignnone size-full wp-image-120416″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q13.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q13″ width=”566″ height=”34″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q13.png 566w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q13-300×18.png 300w” sizes=”(max-width: 566px) 100vw, 566px”> 
Ans: 
Let a and b be the first term and the common difference of the A.P., respectively 
am = a + (m – 1) d = 164 ……………(i) 
Sum of n terms 
Here, n2<script type=”math/tex” id=”MathJax-Element-75″>\frac{n}{2}</script> [2a + nd – d] = 3n2 + 5n 
na+n2d2−nd2=3n2+5n<script type=”math/tex” id=”MathJax-Element-76″>n a+\frac{n^{2} d}{2}-\frac{n d}{2}=3 n^{2}+5 n</script>
n2d2+n(a−d2)<script type=”math/tex” id=”MathJax-Element-77″>\frac{n^{2} d}{2}+n\left(a-\frac{d}{2}\right)</script> = 3n2 + 5n
Comparing the coefficient of n2 on both sides, we obtain 
d2<script type=”math/tex” id=”MathJax-Element-78″>\frac{d}{2}</script> = 3 
⇒ d2 = 2 × 3 = 6 
Comparing the coefficient of n on both sides, we obtain 
a – d2<script type=”math/tex” id=”MathJax-Element-79″>\frac{d}{2}</script> = 5 
a – 62<script type=”math/tex” id=”MathJax-Element-80″>\frac{6}{2}</script> = 5 
a = 5 + 3 = 8 
Therefore, from eq. (i), we obtain 
8 + (m – 1) 6 = 164 
⇒ (m – 1) 6 = 164 – 8 = 156 
⇒ m – 1 = 26 
⇒ m = 27 
Thus the value of m is 27.
Ex 9.2 Class 11 Maths Question 14: 
<img loading=”lazy” class=”alignnone size-full wp-image-120417″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q14.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q14″ width=”528″ height=”29″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q14.png 528w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q14-300×16.png 300w” sizes=”(max-width: 528px) 100vw, 528px”> 
Ans: 
Let A1, A2, A3, A4 and A5 be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 is an A.P. 
Here, a = 8, b = 26, n = 7 
Therefore, 26 = 8 + (7 – 1)d 
6d = 26 – 8 = 18 
d = 3 
A1 = a + d = 8 + 3 = 11 
A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14 
A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17 
A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20 
A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23. 
Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20 and 23.
Ex 9.2 Class 11 Maths Question 15: 
If an+bnan−1+bn−1<script type=”math/tex” id=”MathJax-Element-81″>\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}</script> is the A.M. etween a and b, then find the value of n. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120418″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q15.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q15″ width=”288″ height=”407″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q15.png 288w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q15-212×300.png 212w” sizes=”(max-width: 288px) 100vw, 288px”>
Ex 9.2 Class 11 Maths Question 16: 
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m. 
Ans: 
Let A1, A2, A3, A4, … Am be m A.Ms.between 1 and 31. 
Therefore, 1, A1, A2, A3, ……… Am, 31 are in A.P. 
Let d be the common difference of AP. 
Here, the total number of terms is m + 2 and Tm + 2 = 31 
⇒ 1 + (m + 2 – 1) d = 31 
⇒ (m + 1) d = 30 
⇒ d = 30m+1<script type=”math/tex” id=”MathJax-Element-82″>\frac{30}{m + 1}</script> ……………….(i) 
A7 = T8 = a + 7d 
<img loading=”lazy” class=”alignnone size-full wp-image-138825″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-4.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 4″ width=”560″ height=”330″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-4.png 560w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-4-300×177.png 300w” sizes=”(max-width: 560px) 100vw, 560px”> 
m+21131m−29=59<script type=”math/tex” id=”MathJax-Element-83″>\frac{m+211}{31 m-29}=\frac{5}{9}</script> 
⇒ 9m + 1899 = 155m – 145 
⇒ 146m = 2044 
⇒ m = 2044146<script type=”math/tex” id=”MathJax-Element-84″>\frac{2044}{146}</script> = 14 
∴ m = 14
Ex 9.2 Class 11 Maths Question 17: 
<img loading=”lazy” class=”alignnone size-full wp-image-120421″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q17.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q17″ width=”584″ height=”40″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q17.png 584w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q17-300×21.png 300w” sizes=”(max-width: 584px) 100vw, 584px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120420″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q17.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2 Q17.1″ width=”472″ height=”407″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q17.1.png 472w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.2-Q17.1-300×259.png 300w” sizes=”(max-width: 472px) 100vw, 472px”>
Ex 9.2 Class 11 Maths Question 18: 
The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon. 
Ans: 
The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°. 
It is known that the sum of all angles of a polygon with n sides is 180° (n – 2). 
∵ Sn = 180° (n – 2) 
n2<script type=”math/tex” id=”MathJax-Element-85″>\frac{n}{2}</script> [2a + (n -1) d] = 180° (n – 2) 
n2<script type=”math/tex” id=”MathJax-Element-86″>\frac{n}{2}</script> [240° + (n – 1) 5°] = 180 (n – 2) 
n [240 + (n – 1) 5] = 360 (n – 2) 
240n + 5n2 – 5n = 360n – 720 
⇒ 5n2 + 235n – 360n + 720 = 0 
⇒ 5n2 – 125n + 720 = 0 
⇒ n2 – 25n + 144 = 0 
⇒ n2 – 16n – 9n + 144 = 0 
⇒ n (n – 16) – 9 (n – 16) = 0 
⇒ (n – 9) (n – 16) = 0 
⇒ n = 9 or 16.
<h3 style=”text-align: center;”>NCERT Solutions for class 11 Maths Chapter 9 Exercise 9.3</h3>
Ex 9.3 Class 11 Maths Question 1: 
<img loading=”lazy” class=”alignnone size-full wp-image-120423″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q1″ width=”321″ height=”49″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q1.png 321w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q1-300×46.png 300w” sizes=”(max-width: 321px) 100vw, 321px”> 
Ans: 
The given G.P. is 52,54,58,…<script type=”math/tex” id=”MathJax-Element-87″>\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots</script>. 
Here, a = first term = 52<script type=”math/tex” id=”MathJax-Element-88″>\frac{5}{2}</script> 
r = Common ratio = 545=12<script type=”math/tex” id=”MathJax-Element-89″>\frac{\frac{5}{4}}{5}=\frac{1}{2}</script>
a20 = ar20 – 1
= 52(12)19<script type=”math/tex” id=”MathJax-Element-90″>\frac{5}{2}\left(\frac{1}{2}\right)^{19}</script>
= 5(2)(2)19=5(2)20<script type=”math/tex” id=”MathJax-Element-91″>\frac{5}{(2)(2)^{19}}=\frac{5}{(2)^{20}}</script>
an = arn – 1
= 52(12)n−1<script type=”math/tex” id=”MathJax-Element-92″>\frac{5}{2}\left(\frac{1}{2}\right)^{n-1}</script>
= 5(2)(2)n−1=5(2)n<script type=”math/tex” id=”MathJax-Element-93″>\frac{5}{(2)(2)^{n-1}}=\frac{5}{(2)^{n}}</script>
Ex 9.3 Class 11 Maths Question 2: 
<img loading=”lazy” class=”alignnone size-full wp-image-120425″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q2″ width=”499″ height=”31″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q2.png 499w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q2-300×19.png 300w” sizes=”(max-width: 499px) 100vw, 499px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120424″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q2.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q2.1″ width=”327″ height=”343″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q2.1.png 327w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q2.1-286×300.png 286w” sizes=”(max-width: 327px) 100vw, 327px”>
Ex 9.3 Class 11 Maths Question 3: 
<img loading=”lazy” class=”alignnone size-full wp-image-120427″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q3.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q3″ width=”538″ height=”35″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q3.png 538w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q3-300×20.png 300w” sizes=”(max-width: 538px) 100vw, 538px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120426″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q3.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q3.1″ width=”475″ height=”521″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q3.1.png 475w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q3.1-274×300.png 274w” sizes=”(max-width: 475px) 100vw, 475px”>
Ex 9.3 Class 11 Maths Question 4: 
<img loading=”lazy” class=”alignnone size-full wp-image-120428″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q4.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q4″ width=”576″ height=”38″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q4.png 576w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q4-300×20.png 300w” sizes=”(max-width: 576px) 100vw, 576px”> 
Ans: 
Let a be the first term and r be the common ratio of the G.P. 
∴ a = – 3 
It is known that, 
an = a rn – 1 
∴ a4 = ar3 
= (- 3) r3 
a2 = ar1 
= (- 3) r 
According to the given condition. 
(- 3) r3 = [(- 3) r]2 
⇒ – 3r3 = 9r2 
⇒ r = – 3 
a7 = ar7 – 1 = ar6 
= (- 3) (- 3)6 
= – (3)7 = – 2187 
Thus, the seventh term of the G.P. is – 2187.
Ex 9.3 Class 11 Maths Question 5: 
Which term of the following sequences : 
(a) 2, 2√2, 4, ……….. is 128 ? 
(b) √3, 3, 3√3, ………… is 729? 
(c) 13,19,127,…<script type=”math/tex” id=”MathJax-Element-94″>\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots</script> is 119683<script type=”math/tex” id=”MathJax-Element-95″>\frac{1}{19683}</script>? 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120429″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q5.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q5.1″ width=”416″ height=”463″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q5.1.png 416w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q5.1-270×300.png 270w” sizes=”(max-width: 416px) 100vw, 416px”> 
(b) 
<img loading=”lazy” class=”alignnone size-full wp-image-139114″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-5.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 5″ width=”448″ height=”279″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-5.png 448w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-5-300×187.png 300w” sizes=”(max-width: 448px) 100vw, 448px”> 
<img loading=”lazy” class=”alignnone size-full wp-image-120430″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q5.3.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q5.3″ width=”400″ height=”435″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q5.3.png 400w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q5.3-276×300.png 276w” sizes=”(max-width: 400px) 100vw, 400px”>
Ex 9.3 Class 11 Maths Question 6: 
For what values of x, the numbers – 27<script type=”math/tex” id=”MathJax-Element-96″>\frac{2}{7}</script>, x, – 72<script type=”math/tex” id=”MathJax-Element-97″>\frac{7}{2}</script> are in GP.? 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120431″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q6.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q6.1″ width=”396″ height=”391″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q6.1.png 396w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q6.1-300×296.png 300w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q6.1-100×100.png 100w” sizes=”(max-width: 396px) 100vw, 396px”>
Ex 9.3 Class 11 Maths Question 7: 
Find the sum to 20 terms In the geometric progression 0.15, 0.015, 0.0015 ……… 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120432″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q7.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q7.1″ width=”314″ height=”337″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q7.1.png 314w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q7.1-280×300.png 280w” sizes=”(max-width: 314px) 100vw, 314px”>
Ex 9.3 Class 11 Maths Question 8: 
<img loading=”lazy” class=”alignnone size-full wp-image-120433″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q8.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q8″ width=”465″ height=”40″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q8.png 465w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q8-300×26.png 300w” sizes=”(max-width: 465px) 100vw, 465px”> 
Ans: 
The given G.P. is √7, √21, 3√7 …………… 
Here, a = √7 
<img loading=”lazy” class=”alignnone size-full wp-image-139115″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-6.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 6″ width=”517″ height=”342″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-6.png 517w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-6-300×198.png 300w” sizes=”(max-width: 517px) 100vw, 517px”>
Ex 9.3 Class 11 Maths Question 9: 
Find the sum to n terms in the geometric progression 1, – a, a2, – a3 ………… (if a ≠ 1) 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120434″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q9.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q9.1″ width=”286″ height=”234″>
Ex 9.3 Class 11 Maths Question 10: 
Find the sum to n terms in the geometric progression x3, x5, x7 … (if x ≠ ± 1). 
Ans: 
The given G.P. is x3, x5, x7, …………. 
Here, a = x3 and r = x2 
Sn = a(1−rn)1−r<script type=”math/tex” id=”MathJax-Element-98″>\frac{a\left(1-r^{n}\right)}{1-r}</script>
= x3[1−(x2)n]1−x2=x3(1−x2n)1−x2<script type=”math/tex” id=”MathJax-Element-99″>\frac{x^{3}\left[1-\left(x^{2}\right)^{n}\right]}{1-x^{2}}=\frac{x^{3}\left(1-x^{2 n}\right)}{1-x^{2}}</script>.
Ex 9.3 Class 11 Maths Question 11: 
Evaluate ∑−k=111<script type=”math/tex” id=”MathJax-Element-100″>\sum-{k=1}^{11}</script> (2 + 3k) 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-139116″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-7.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 7″ width=”572″ height=”386″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-7.png 572w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-7-300×202.png 300w” sizes=”(max-width: 572px) 100vw, 572px”>
Ex 9.3 Class 11 Maths Question 12: 
The sum of first three terms of a G.P. is – and their product is 1. Find the common ratio and the terms. 
Ans: 
Let the first three number of G.P. be ar<script type=”math/tex” id=”MathJax-Element-101″>\frac{a}{r}</script>, a and ar. 
According to the question, 
ar<script type=”math/tex” id=”MathJax-Element-102″>\frac{a}{r}</script> + a + ar = 3910<script type=”math/tex” id=”MathJax-Element-103″>\frac{39}{10}</script> … (i) 
and (ar<script type=”math/tex” id=”MathJax-Element-104″>\frac{a}{r}</script>) × (a) × (ar) = 1 
⇒ a3 = 1 
⇒ a = 1 
On putting the value of a = 1 in eq. (i), we get 
1r<script type=”math/tex” id=”MathJax-Element-105″>\frac{1}{r}</script> + 1 + r = 3910<script type=”math/tex” id=”MathJax-Element-106″>\frac{39}{10}</script> 
1+r+r2r=3910<script type=”math/tex” id=”MathJax-Element-107″>\frac{1+r+r^{2}}{r}=\frac{39}{10}</script> 
⇒ 10 + 10r + 10r2 = 39r 
⇒ 10r2 + 10r – 39r + 10 = 0 
⇒ 10r2 – 29r + 10 = 0 
Now, factorising it by splitting the middle term, we get 
10r2 – 25r – 4r + 10 = 0 
⇒ 5r (2r – 5) – 2 (2r – 5) = 0 
⇒ (5r – 2)(2r – 5) = 0 
⇒ 5r – 2 = 0 and 2r – 5 = 0 
⇒ r = 25<script type=”math/tex” id=”MathJax-Element-108″>\frac{2}{5}</script> and r = 52<script type=”math/tex” id=”MathJax-Element-109″>\frac{5}{2}</script> 
When a = 1 and r = 25<script type=”math/tex” id=”MathJax-Element-110″>\frac{2}{5}</script>, then numbers are 
ar=125=52<script type=”math/tex” id=”MathJax-Element-111″>\frac{a}{r}=\frac{1}{\frac{2}{5}}=\frac{5}{2}</script>, 
a = 1 and 
ar = 1 × 25<script type=”math/tex” id=”MathJax-Element-112″>\frac{2}{5}</script> = 25<script type=”math/tex” id=”MathJax-Element-113″>\frac{2}{5}</script> 
∴ 52<script type=”math/tex” id=”MathJax-Element-114″>\frac{5}{2}</script>, 1, 25<script type=”math/tex” id=”MathJax-Element-115″>\frac{2}{5}</script>.
When a = 1 and r = 52<script type=”math/tex” id=”MathJax-Element-116″>\frac{5}{2}</script>, then numbers are 
ar=15=25<script type=”math/tex” id=”MathJax-Element-117″>\frac{a}{r}=\frac{1}{5}=\frac{2}{5}</script>;
a = 1 and ar = 1 × 52<script type=”math/tex” id=”MathJax-Element-118″>\frac{5}{2}</script> = 52<script type=”math/tex” id=”MathJax-Element-119″>\frac{5}{2}</script> 
∴ 25<script type=”math/tex” id=”MathJax-Element-120″>\frac{2}{5}</script>, 1, 52<script type=”math/tex” id=”MathJax-Element-121″>\frac{5}{2}</script>.
Ex 9.3 Class 11 Maths Question 13: 
How many terms of G.P. 3, 32, 33, …………… are needed to give the sum 120? 
Ans: 
The given G.P. is 3, 32, 33, ………… 
Let n terms of this G.P. be required to obtain the sum as 120. 
Sn = a(rn−1)r−1<script type=”math/tex” id=”MathJax-Element-122″>\frac{a\left(r^{n}-1\right)}{r-1}</script> 
Here, a = 3 and r = 3 
Sn = 120 = 3(3n−1)3−1<script type=”math/tex” id=”MathJax-Element-123″>\frac{3\left(3^{n}-1\right)}{3-1}</script>
⇒ 120 = 3(3n−1)3−1<script type=”math/tex” id=”MathJax-Element-124″>\frac{3\left(3^{n}-1\right)}{3-1}</script> 
⇒ 120×23<script type=”math/tex” id=”MathJax-Element-125″>\frac{120 \times 2}{3}</script> = 3n – 1 
⇒ 3n – 1 = 80 
⇒ 3n = 81 
⇒ 3n = 34 
∴ n = 4 
Thus, four terms of the given G.P. are required to obtain the sum as 120.
Ex 9.3 Class 11 Maths Question 14: 
The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P. 
Ans: 
Let a be the first term and r be the common ratio, then 
<img loading=”lazy” class=”alignnone size-full wp-image-139117″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-8.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 8″ width=”560″ height=”397″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-8.png 560w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-8-300×213.png 300w” sizes=”(max-width: 560px) 100vw, 560px”><img loading=”lazy” class=”alignnone size-full wp-image-139120″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-10.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 10″ width=”422″ height=”63″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-10.png 422w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-10-300×45.png 300w” sizes=”(max-width: 422px) 100vw, 422px”>
Ex 9.3 Class 11 Maths Question 15: 
Given a G.P. with a = 729 and 7th term 64, determine S7. 
Ans: 
a = 729, a7 = 64 
Let r be the common ratio of the G.P. 
It is known that, an = arn – 1 
a7 = ar7 – 1 (729)r6 
<img loading=”lazy” class=”alignnone size-full wp-image-139121″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-11.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 11″ width=”467″ height=”337″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-11.png 467w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-11-300×216.png 300w” sizes=”(max-width: 467px) 100vw, 467px”>
Ex 9.3 Class 11 Maths Question 16: 
Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term. 
Ans: 
Let, first term = a and common ratio = r. 
Given, sum of first two terms = – 4 
⇒ a1 + a2 = – 4 
⇒ a + ar = – 4 
⇒ a (1 + r) = – 4 
and fifth term = 4 × third term 
a5 = 4 × a3 
⇒ ar4 = 4ar2 
r2 = 4 
r = ± 2 
When r = 2, then from eq. (i), we get 
a (1 + 2) = – 4 
⇒ a = – 43<script type=”math/tex” id=”MathJax-Element-126″>\frac{4}{3}</script> 
Then, G.P. is – 43<script type=”math/tex” id=”MathJax-Element-127″>\frac{4}{3}</script>, – 43<script type=”math/tex” id=”MathJax-Element-128″>\frac{4}{3}</script> × 2, – 43<script type=”math/tex” id=”MathJax-Element-129″>\frac{4}{3}</script> × (2)2, …………… 
i.e., −43,−83,−163<script type=”math/tex” id=”MathJax-Element-130″>\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}</script> 
When r = – 2, then from eq. (i), we get a 
a (1 – 2)= – 4 
⇒ – a = – 4 
⇒ a = 4 
Then, G.P. is 4, 4 × (- 2), 4 × (- 2)2 … i.e, 4,- 8, 16, ………..
Ex 9.3 Class 11 Maths Question 17: 
If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P. 
Ans: 
Given, 4 th term, 
T4 = x 
⇒ ar4 – 1 = x 
⇒ ar3 = x …………………..(i) 
10th term, 
T10 = y 
⇒ ar10 – 1 = y 
⇒ ar9 = y ………………….(ii) 
and 16 th term, T16 = z 
⇒ ar16 – 1 = z 
⇒ ar15 = z ………………(iii) 
Now, multiplying eq. (j) by eq. (ill), we get 
ar3 × ar15 = x × z 
a2 r3 + 15 = x × z 
a2 r18 = xz 
(ar9)2 = xz 
∴ y2 = xz [from eq. (ii)] 
Therefore, x,y and z are in GP.
Ex 9.3 Class 11 Maths Question 18: 
<img loading=”lazy” class=”alignnone size-full wp-image-120442″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q18.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q18″ width=”412″ height=”36″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q18.png 412w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q18-300×26.png 300w” sizes=”(max-width: 412px) 100vw, 412px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120441″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q18.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q18.1″ width=”579″ height=”432″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q18.1.png 579w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q18.1-300×224.png 300w” sizes=”(max-width: 579px) 100vw, 579px”>
Ex 9.3 Class 11 Maths Question 19: 
<img loading=”lazy” class=”alignnone size-full wp-image-120443″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q19.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q19″ width=”572″ height=”67″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q19.png 572w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q19-300×35.png 300w” sizes=”(max-width: 572px) 100vw, 572px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-139161″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-12.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 12″ width=”524″ height=”406″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-12.png 524w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-12-300×232.png 300w” sizes=”(max-width: 524px) 100vw, 524px”>
Ex 9.3 Class 11 Maths Question 20: 
Show that the products of the corresponding terms of the sequences a, ar, ar2, … arn – 1 and A, AR, AR2, …….. ARn – 1 form a G.P. and find the common ratio. 
Ans: 
It has to be proved that the sequence, aA, arAR, ar2AR2, ………. arn – 1 ARn – 1, forms a G.P. 
 Second term  First term =arARaA<script type=”math/tex” id=”MathJax-Element-131″>\frac{\text { Second term }}{\text { First term }}=\frac{a r A R}{a A}</script> = rR
 Third term  Second term =ar2AR2arAR<script type=”math/tex” id=”MathJax-Element-132″>\frac{\text { Third term }}{\text { Second term }}=\frac{a r^{2} A R^{2}}{a r A R}</script> = rR 
Thus, the above sequence forms a G.P. and the common ratio is rR.
Ex 9.3 Class 11 Maths Question 21: 
Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18. 
Ans: 
Let a be the first term and r the common ratio of G.P. 
∴ nth term = Tn = arn – 1 
⇒ T2 = ar , T3 = ar2 and T4 = ar3 
Since third term is greater than the first by 9. 
∴ T3 = T1 + 9 
⇒ ar2 = a + 9 
Second term is greater than the 4th by 18. 
T2 = T4 + 18 
⇒ ar = ar3 + 18 
⇒ ar3 = ar + 9r 
From eqs. (ii) and (iii), we get 
ar = ar + 9r + 18 
⇒ 0 = 9r + 18 
⇒ r = −189<script type=”math/tex” id=”MathJax-Element-133″>\frac{-18}{9}</script> = – 2 
Put = – 2 in (i), we get 
a(- 2)2 = a + 9 
⇒ 4a = a + 9 
⇒ 3a = 9 
⇒ a = 3 
T2 = ar = 3 (- 2) = – 6 
T3 = ar2 
= 3 (- 2)2 = 12 
T4 = ar3 
= 3 (- 2)3 = – 24 
∴ Required terms are 3, – 6, 12 and – 24.
Ex 9.3 Class 11 Maths Question 22: 
<img loading=”lazy” class=”alignnone size-full wp-image-120447″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q22.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q22″ width=”590″ height=”38″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q22.png 590w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q22-300×19.png 300w” sizes=”(max-width: 590px) 100vw, 590px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120446″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q22.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q22.1″ width=”488″ height=”417″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q22.1.png 488w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q22.1-300×256.png 300w” sizes=”(max-width: 488px) 100vw, 488px”>
Ex 9.3 Class 11 Maths Question 23: 
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n. 
Ans: 
The first term of the G.P. is a and the last term is b. 
Therefore, the G.P. is a, ar, ar2, ar3, …………. arn – 1, where r is the common ratio. 
b = arn – 1 …………..(i) 
P = Product of n terms 
= (a) (ar) (ar2) … (arn – 1) 
= (a × a × ……… a) (r × r2 × ……….. rn – 1) 
= an r1 + 2 + ………. + (n – 1) 
Here, 1, 2, ………. (n – 1) is an A.P. 
<img loading=”lazy” class=”alignnone size-full wp-image-139182″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-13.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 13″ width=”546″ height=”180″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-13.png 546w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-13-300×99.png 300w” sizes=”(max-width: 546px) 100vw, 546px”> 
Thus, the given result is proved.
Ex 9.3 Class 11 Maths Question 24: 
<img loading=”lazy” class=”alignnone size-full wp-image-120449″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q24.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q24″ width=”500″ height=”71″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q24.png 500w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q24-300×43.png 300w” sizes=”(max-width: 500px) 100vw, 500px”> 
Ans: 
Let a be the first term and r be the common ratio of the G.P. 
Sum of first n terms = a(1−rn)(1−r)<script type=”math/tex” id=”MathJax-Element-134″>\frac{a\left(1-r^{n}\right)}{(1-r)}</script> 
Since there are n terms from (n + 1)th to (2n)th term, 
sum of terms from (n + 1)th to (2n)th term 
= an+1(1−rn)(1−r)<script type=”math/tex” id=”MathJax-Element-135″>\frac{a_{n+1}\left(1-r^{n}\right)}{(1-r)}</script> 
= arn(1−rn)(1−r)<script type=”math/tex” id=”MathJax-Element-136″>\frac{a r^{n}\left(1-r^{n}\right)}{(1-r)}</script> [∵ an + 1 = arn + 1 – 1 = arn]
Thus, required ratio = a(1−rn)(1−r)×(1−r)arn(1−rn)=1rn<script type=”math/tex” id=”MathJax-Element-137″>\frac{a\left(1-r^{n}\right)}{(1-r)} \times \frac{(1-r)}{a r^{n}\left(1-r^{n}\right)}=\frac{1}{r^{n}}</script>
Thus, the ratio of the sum of first n terms of G.P. to the sum of terms from (n + 1)th to (2n)th term is 1rn<script type=”math/tex” id=”MathJax-Element-138″>\frac{1}{r^{n}}</script>.
Ex 9.3 Class 11 Maths Question 25: 
<img loading=”lazy” class=”alignnone size-full wp-image-120452″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q25.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q25″ width=”539″ height=”40″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q25.png 539w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q25-300×22.png 300w” sizes=”(max-width: 539px) 100vw, 539px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120450″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q25.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q25.1″ width=”354″ height=”485″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q25.1.png 354w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q25.1-219×300.png 219w” sizes=”(max-width: 354px) 100vw, 354px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120451″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q25.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q25.2″ width=”512″ height=”352″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q25.2.png 512w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q25.2-300×206.png 300w” sizes=”(max-width: 512px) 100vw, 512px”>
Ex 9.3 Class 11 Maths Question 26: 
<img loading=”lazy” class=”alignnone size-full wp-image-120454″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q26.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q26″ width=”487″ height=”35″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q26.png 487w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q26-300×22.png 300w” sizes=”(max-width: 487px) 100vw, 487px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120453″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q26.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q26.1″ width=”550″ height=”347″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q26.1.png 550w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q26.1-300×189.png 300w” sizes=”(max-width: 550px) 100vw, 550px”>
Ex 9.3 Class 11 Maths Question 27: 
Find the value of n so that an+1+bn+1an+bn<script type=”math/tex” id=”MathJax-Element-139″>\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}</script> may be the geometric mean between a and b. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-139119″ src=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-14.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 14″ width=”570″ height=”537″ srcset=”https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-14.png 570w, https://www.learncbse.in/wp-content/uploads/2019/09/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-14-300×283.png 300w” sizes=”(max-width: 570px) 100vw, 570px”> 
Hence proved.
Ex 9.3 Class 11 Maths Question 28: 
<img loading=”lazy” class=”alignnone size-full wp-image-120457″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q28.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q28″ width=”570″ height=”58″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q28.png 570w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q28-300×31.png 300w” sizes=”(max-width: 570px) 100vw, 570px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120455″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q28.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q28.1″ width=”316″ height=”454″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q28.1.png 316w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q28.1-209×300.png 209w” sizes=”(max-width: 316px) 100vw, 316px”> 
<img loading=”lazy” class=”alignnone size-full wp-image-120456″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q28.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q28.2″ width=”352″ height=”230″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q28.2.png 352w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q28.2-300×196.png 300w” sizes=”(max-width: 352px) 100vw, 352px”>
Ex 9.3 Class 11 Maths Question 29: 
If A and G be AM. and G.M., respectively between two positive numbers, prove that the numbers are A ± (A+G)(A−G)−−−−−−−−−−−−−√<script type=”math/tex” id=”MathJax-Element-140″>\sqrt{(A+G)(A-G)}</script>. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120458″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q29.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q29.1″ width=”584″ height=”432″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q29.1.png 584w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q29.1-300×222.png 300w” sizes=”(max-width: 584px) 100vw, 584px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120459″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q29.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q29.2″ width=”395″ height=”243″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q29.2.png 395w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q29.2-300×185.png 300w” sizes=”(max-width: 395px) 100vw, 395px”>
Ex 9.3 Class 11 Maths Question 30: 
<img loading=”lazy” class=”alignnone size-full wp-image-120460″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q30.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q30″ width=”583″ height=”65″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q30.png 583w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q30-300×33.png 300w” sizes=”(max-width: 583px) 100vw, 583px”> 
Ans: 
It is given that the number of bacteria doubles every hour. 
Therefore, the number of bacteria after every hour will form a G.P. 
Here, a = 30 and r = 2 
∴ a3 = ar2 
= (30) (2)2 = 120 
Therefore, the number of bacteria at the end of 2nd hour will be 120. 
a5 = ar4 
= (30) (2)4 = 480 
The number of bacteria at the end of 4th hour will be 480. 
an + 1 = arn = (30) 2n 
Thus, number of bacteria at the end of nth hour will be 30(2)n.
Ex 9.3 Class 11 Maths Question 31: 
<img loading=”lazy” class=”alignnone size-full wp-image-120461″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q31.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q31″ width=”566″ height=”48″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q31.png 566w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q31-300×25.png 300w” sizes=”(max-width: 566px) 100vw, 566px”> 
Ans: 
The amount deposited in the bank is Rs. 500. 
At the end of first year, amount = Rs. 500(1 + 110<script type=”math/tex” id=”MathJax-Element-141″>\frac{1}{10}</script>) = Rs. 500 (1.1) 
At the end of 2nd year, amount = Rs. 500 (1.1) (1.1) 
At the end of 3rd year, amount = Rs. 500 (1.1) (1.1) (1.1) and so on . 
Amount at the end of 10 years = Rs. 500 (1.1) (1.1) … (10 times) 
= Rs. 500 (1.1)10.
Ex 9.3 Class 11 Maths Question 32: 
If AM. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120462″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q32.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3 Q32.1″ width=”452″ height=”348″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q32.1.png 452w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.3-Q32.1-300×231.png 300w” sizes=”(max-width: 452px) 100vw, 452px”>
<h3 style=”text-align: center;”>NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.4</h3>
Ex 9.4 Class 11 Maths Question 1: 
Find the sum to n terms of the series 
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …………… 
Ans: 
Let S = 1 . 2 + 2 . 3 + 3 . 4 + 4 . 5 + ……………. 
Then, nth term, 
Tn = n(n + 1) = n2 + n 
∴ Tn = n2 + n 
On taking summation from 1 to n on both sides we get 
<img loading=”lazy” class=”alignnone size-full wp-image-138919″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 1″ width=”437″ height=”150″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-1.png 437w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-1-300×103.png 300w” sizes=”(max-width: 437px) 100vw, 437px”>
Ex 9.4 Class 11 Maths Question 2: 
Find the sum to n terms of the series 
1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + ………….. 
Ans: 
The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + ………….. 
nth term an = n (n + 1) (n + 2) 
= (n2 + n) (n + 2) = n2 + 3n2 + 2n 
<img loading=”lazy” class=”alignnone size-full wp-image-138921″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 2″ width=”526″ height=”329″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-2.png 526w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-2-300×188.png 300w” sizes=”(max-width: 526px) 100vw, 526px”>
Ex 9.4 Class 11 Maths Question 3: 
Find the sum of n terms of the series 3 × 12 + 5 × 22 + 7 × 32 + …………….. 
Ans:
The given series is 3 × 12 + 5 × 22 + 7 × 32 + …………….. 
nth term an = (2n + 1) n2 
= 2n3 + n2 
∴ Sn = ∑nk=1ak=∑nk=1(2k3+k2)=2∑nk=1k3+∑nk=1k2<script type=”math/tex” id=”MathJax-Element-142″>\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(2 k^{3}+k^{2}\right)=2 \sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n} k^{2}</script>
= 2[n(n+1)2]2+n(n+1)(2n+1)6<script type=”math/tex” id=”MathJax-Element-143″>2\left[\frac{n(n+1)}{2}\right]^{2}+\frac{n(n+1)(2 n+1)}{6}</script> 
<img loading=”lazy” class=”alignnone size-full wp-image-138928″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-3.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 3″ width=”596″ height=”174″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-3.png 596w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-3-300×88.png 300w” sizes=”(max-width: 596px) 100vw, 596px”>
Ex 9.4 Class 11 Maths Question 4: 
Find the sum to n terms of the series 11×2+12×3+13×4<script type=”math/tex” id=”MathJax-Element-144″>\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}</script> + ……. 
Ans: 
Let the given series be 
S = 11×2+12×3+13×4<script type=”math/tex” id=”MathJax-Element-145″>\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}</script> + ……. 
Then, nth term Tn = 1n(n+1)<script type=”math/tex” id=”MathJax-Element-146″>\frac{1}{n(n+1)}</script> 
Now, we will split the denominator of the nth term into two parts or we will write Tn as the difference of two terms. 
<img loading=”lazy” class=”alignnone size-full wp-image-138935″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-4.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 4″ width=”463″ height=”337″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-4.png 463w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-4-300×218.png 300w” sizes=”(max-width: 463px) 100vw, 463px”> <img loading=”lazy” class=”alignnone size-full wp-image-138936″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-5.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 5″ width=”505″ height=”281″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-5.png 505w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-5-300×167.png 300w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-5-360×200.png 360w” sizes=”(max-width: 505px) 100vw, 505px”> 
Ex 9.4 Class 11 Maths Question 5: 
Find the sum to n terms of the series 52 + 62 + 72 + … + 202. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120463″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q5.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 Q5.1″ width=”357″ height=”491″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q5.1.png 357w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q5.1-218×300.png 218w” sizes=”(max-width: 357px) 100vw, 357px”>
Ex 9.4 Class 11 Maths Question 6: 
Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 + ………….. 
Ans: 
The given series is 3 × 8 + 6 × 11 + 9 × 14 + …………. 
an = (nth term of 3, 6, 9 ………..) × (nth term of 8, 11, 14, …………) 
= (3n) (3n + 5) 
= 9n2 + 15n 
<img loading=”lazy” class=”alignnone size-full wp-image-138943″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-6.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 6″ width=”515″ height=”189″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-6.png 515w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-6-300×110.png 300w” sizes=”(max-width: 515px) 100vw, 515px”>
Ex 9.4 Class 11 Maths Question 7: 
Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …………. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120464″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q7.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 Q7.1″ width=”397″ height=”215″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q7.1.png 397w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q7.1-300×162.png 300w” sizes=”(max-width: 397px) 100vw, 397px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120465″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q7.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 Q7.2″ width=”338″ height=”513″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q7.2.png 338w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q7.2-198×300.png 198w” sizes=”(max-width: 338px) 100vw, 338px”>
Ex 9.4 Class 11 Maths Question 8: 
Find the sum to re terms of the series whose nth term is given by n (n + 1) (n + 4). 
Ans: 
an = n (n + 1) (n + 4) 
= n(n2 + 5n + 4) 
= n3 + 5n2 + 4n 
<img loading=”lazy” class=”alignnone size-full wp-image-138948″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-7.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 7″ width=”393″ height=”327″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-7.png 393w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-7-300×250.png 300w” sizes=”(max-width: 393px) 100vw, 393px”> 
Ex 9.4 Class 11 Maths Question 9: 
Find the sum to nth terms of the series whose nth term is given by the n2 + 2n. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120466″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q9.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 Q9.1″ width=”578″ height=”393″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q9.1.png 578w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-Q9.1-300×204.png 300w” sizes=”(max-width: 578px) 100vw, 578px”>
Ex 9.4 Class 11 Maths Question 10: 
Find the sum to re terms of the series whose nth term is given by (2n – 1)2. 
Ans: 
Given, nth term Tn = (2n – 1)2 
⇒ Tn = 4 n2 + 1 – 4n 
Now, S = Σ Tn 
= Σ (4n2 + 1 – 4n) 
= 4 Σn2 + Σ 1 – 4 Σn 
<img loading=”lazy” class=”alignnone size-full wp-image-138952″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-8.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 8″ width=”516″ height=”297″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-8.png 516w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Ex-9.4-8-300×173.png 300w” sizes=”(max-width: 516px) 100vw, 516px”>
<h3 style=”text-align: center;”>Class 11 Maths NCERT Miscellaneous Solutions</h3>
Miscellaneous Exercise Class 11 Maths Question 1: 
Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120468″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q1.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q1.1″ width=”573″ height=”518″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q1.1.png 573w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q1.1-300×271.png 300w” sizes=”(max-width: 573px) 100vw, 573px”>
Miscellaneous Exercise Class 11 Maths Question 2: 
If the sum of three numbers in AP is 24 and their product is 440, find the numbers. 
Ans: 
Let the three numbers in A.P. be a – d, a, and a + d. 
According to the given information, 
(a – d) + (a) + (a + d) = 24 
3a = 24 
⇒ a = 8 
and (a – d) a (a + d) = 440 ………………(ii) 
(8 – d) (8) (8 + d) = 440 
(8 – d) (8 + d) = 55 
= 64 – d2 = 55 
d2 = 64 – 55 = 9 
d = ±3 
Therefore, when d = 3, the numbers are 5, 8 and 11 and when d = – 3, the numbers are 11, 8 and 5. 
Thus, the three numbers are 5, 8 and 11.
Miscellaneous Exercise Class 11 Maths Question 3: 
Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2 – S1). 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120469″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q3.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q3.1″ width=”570″ height=”520″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q3.1.png 570w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q3.1-300×274.png 300w” sizes=”(max-width: 570px) 100vw, 570px”> 
Hence, S3 = 3 (S2 – S1)
Miscellaneous Exercise Class 11 Maths Question 4: 
Find the sum of all numbers between 200 and 400 which are divisible by 7. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120471″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q4.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q4.1″ width=”551″ height=”408″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q4.1.png 551w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q4.1-300×222.png 300w” sizes=”(max-width: 551px) 100vw, 551px”> 
Sn = n2<script type=”math/tex” id=”MathJax-Element-147″>\frac{n}{2}</script> [2a + (n – 1) d] 
S29 = 292<script type=”math/tex” id=”MathJax-Element-148″>\frac{29}{2}</script> [2 × 203 + (29 – 1) 7] 
= 292<script type=”math/tex” id=”MathJax-Element-149″>\frac{29}{2}</script> [406 + 28 × 7] 
= 292<script type=”math/tex” id=”MathJax-Element-150″>\frac{29}{2}</script> [406 + 196] 
= 292<script type=”math/tex” id=”MathJax-Element-151″>\frac{29}{2}</script> × 602 2 2 2 
= 29 × 301 = 8729.
Miscellaneous Exercise Class 11 Maths Question 5: 
Find the sum of integers from 1 to 100 that are divisible by 2 or 5. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120472″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q5.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q5.1″ width=”591″ height=”517″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q5.1.png 591w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q5.1-300×262.png 300w” sizes=”(max-width: 591px) 100vw, 591px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120473″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q5.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q5.2″ width=”570″ height=”517″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q5.2.png 570w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q5.2-300×272.png 300w” sizes=”(max-width: 570px) 100vw, 570px”>
Miscellaneous Exercise Class 11 Maths Question 6: 
Find the sum of all two digit numbers which when divided by 4, yields 1 as remninder. 
Ans: 
The sum of two digit numbers divisible by 4 yield 1 as remainder is 13 + 17 + 21 + ………… + 97. 
Let the sum be denoted by S and let 97 be the nth term. 
∴ Tn = a + (n – 1) d 
97 = a + (n – 1) d 
= 13 + (n – 1) 4 
⇒ 97 = 13 + 4n – 4 
⇒ 97 – 9 = 4n 
⇒ n = 22 
∴ The sum, Sn = 13 + 17 + 21 + ………….+ 97 
∴ Sn = n2<script type=”math/tex” id=”MathJax-Element-152″>\frac{n}{2}</script> [2a + (n – 1)d] 
= 227<script type=”math/tex” id=”MathJax-Element-153″>\frac{22}{7}</script> [2 × 13 + (22 – 1) × 4] 
= 11 [26 + 21 × 4] 
= 11 [26 + 84] 
= 11 × 110 = 1210
Miscellaneous Exercise Class 11 Maths Question 7: 
<img loading=”lazy” class=”alignnone size-full wp-image-120478″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q7.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q7″ width=”471″ height=”77″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q7.png 471w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q7-300×49.png 300w” sizes=”(max-width: 471px) 100vw, 471px”> 
Ans: 
It is given that, 
f(x + y) = f(x) × f(y) for all x, y ∈ N 
f(1) = 3 
Taking x = y = 1 in eq. (i), we obtain 
f(1 + 1) = f(2) = f(1) 
f(1) = 3 × 3 = 9 
Similarly, f(1 + 1 + 1) = f(3) = f(1 + 2) = f(1) f(2) = 3 × 9 = 27 
f(4) = f(1 + 3) = f(1) f(3) = 3 × 27 = 81 
f(1), f(2), f(3) , that is 3, 9, 27, …………, forms a G.P. with both the first term and common ratio equal to 3. 
It is known that, Sn = a(rn−1)r−1<script type=”math/tex” id=”MathJax-Element-154″>\frac{a\left(r^{n}-1\right)}{r-1}</script> 
It is given that, ∑nx=1<script type=”math/tex” id=”MathJax-Element-155″>\sum_{x=1}^{n}</script> f(x) = 120 
∴ 120 = 3(3n−1)3−1<script type=”math/tex” id=”MathJax-Element-156″>\frac{3\left(3^{n}-1\right)}{3-1}</script> 
<img loading=”lazy” class=”alignnone size-full wp-image-120477″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q7.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q7.2″ width=”199″ height=”148″>
Miscellaneous Exercise Class 11 Maths Question 8: 
The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms. 
Ans: 
Let the sum of n terms of the G.P. be 315. 
It is known that, Sn = a(rn−1)r−1<script type=”math/tex” id=”MathJax-Element-157″>\frac{a\left(r^{n}-1\right)}{r-1}</script> 
It is given that the first term a is 5 and common ratio r is 2. 
315 = 5(2n−1)2−1<script type=”math/tex” id=”MathJax-Element-158″>\frac{5\left(2^{n}-1\right)}{2-1}</script> 
⇒ 2n = 64 = (2)6 
⇒ n = 6 
∴ Last term of the G.P.= 6th term 
= ar6 – 1 = (5) (2)5 = (5) (32) = 160 
Thus, the last term of the G.P. is 160.
Miscellaneous Exercise Class 11 Maths Question 9: 
The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120480″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q9.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q9.1″ width=”490″ height=”368″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q9.1.png 490w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q9.1-300×225.png 300w” sizes=”(max-width: 490px) 100vw, 490px”>
Miscellaneous Exercise Class 11 Maths Question 10: 
The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers. 
Ans: 
Let the three numbers in G.P. be a, ar and ar2. 
From the given condition, a + ar + ar2 = 56 
a (1 + r + r2) = 56 
a = 561+r+r2<script type=”math/tex” id=”MathJax-Element-159″>\frac{56}{1+r+r^{2}}</script> …………………(i) 
a – 1, ar – 7, ar2 – 21 forms an A.P. 
∴ (ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7) 
⇒ ar – a – 6 = ar2 – ar – 14 
⇒ ar2 – 2ar + a = 8 
⇒ ar2 – ar – ar + a = 8 
a(r2 + 1 – 2r) = 8 
a(r – 1)2 = 8 ………….(ii) 
561+r+r2<script type=”math/tex” id=”MathJax-Element-160″>\frac{56}{1+r+r^{2}}</script> (r – 1)2 = 8 [using eq. (1)] 
⇒ 7 (r2 – 2r + 1) = 1 + r + r2 
⇒ 7r2 – 14r + 7 – 1 – r – r2 = 0 
⇒ 6r2 – 15r + 6 = 0 
⇒ r2 – 12r – 3r + 6 = 0 
⇒ 6r (r – 2) – 3 (r – 2) = 0 
⇒ (6r – 3)(r – 2) = 0 
∴ r = 2, 12<script type=”math/tex” id=”MathJax-Element-161″>\frac{1}{2}</script> 
When r = 2, a = 8; When r = 12<script type=”math/tex” id=”MathJax-Element-162″>\frac{1}{2}</script>, a = 32 
Therefore, when r = 2, the three numbers in G.P. are 8, 16 and 32. 
When r = 12<script type=”math/tex” id=”MathJax-Element-163″>\frac{1}{2}</script>, the three numbers in G.P. are 32, 16 and 8. 
Thus, in either case, the three required numbers are 8, 16, and 32.
Miscellaneous Exercise Class 11 Maths Question 11: 
A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120481″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q11.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q11.1″ width=”358″ height=”415″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q11.1.png 358w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q11.1-259×300.png 259w” sizes=”(max-width: 358px) 100vw, 358px”> 
Miscellaneous Exercise Class 11 Maths Question 12: 
The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms. 
Ans: 
Let the A.P. be a, a + d, a + 2d, a + 3d,… a + (n – 2) d, a + (n – 1 )d 
Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d 
Sum of last four terms = [a + (n – 4) d] + [a +(n – 3) d] + [a + (n – 2) d] + [(a + n – 1) d] 
= 4a + (4n – 10)d 
According to the given condition, 
4a + 6d = 56 
⇒ 4 (11) + 6d = 56 [Since a = 11 (given)] 
⇒ 6d = 12 
⇒ d = 2 
4a + (4n – 10) d = 112 
⇒ 4(11) + (4n – 10) 2 = 112 
⇒ (4n – 10)2 = 68 
⇒ 4n – 10 = 34 
4n = 44 
n = 11 
Thus, the number of terms of the A.P. is 11.
Miscellaneous Exercise Class 11 Maths Question 13: 
If a+bxa−bx=b+cxb−cx=c+dxc−dx<script type=”math/tex” id=”MathJax-Element-164″>\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}</script>, then show that a, b, c and d are in G.P. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120483″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q13.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q13.1″ width=”357″ height=”438″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q13.1.png 357w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q13.1-245×300.png 245w” sizes=”(max-width: 357px) 100vw, 357px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120484″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q13.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q13.2″ width=”254″ height=”127″>
Miscellaneous Exercise Class 11 Maths Question 14: 
Let S be the sum, P the product and R the sum of reciprocal of n terms in a G.P. Prove that P2 Rn = Sn. 
Ans: 
Let the G.P. be a, ar, ar2, ar3, ………….., arn – 1 
According to the given information, 
S = a(rn−1)r−1<script type=”math/tex” id=”MathJax-Element-165″>\frac{a\left(r^{n}-1\right)}{r-1}</script> 
P = an rn(n−1)2<script type=”math/tex” id=”MathJax-Element-166″>\frac{n(n-1)}{2}</script> 
[∵ Sum of n natural numbers is n (n+1)2<script type=”math/tex” id=”MathJax-Element-167″>\frac{(n+1)}{2}</script>] 
<img loading=”lazy” class=”alignnone size-full wp-image-139010″ src=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex 1″ width=”424″ height=”289″ srcset=”https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-1.png 424w, https://www.learncbse.in/wp-content/uploads/2022/06/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-1-300×204.png 300w” sizes=”(max-width: 424px) 100vw, 424px”> 
Hence P2 Rn = Sn.
Miscellaneous Exercise Class 11 Maths Question 15: 
The th, qth and rth terms of an AP. are a, b, c respectively. Show that(q – r)a + (r – p) b + (p – q) c = 0. 
Ans: 
Let A be the first term and d be the common difference. 
Since, Tp = a 
⇒ A + (p – 1) d = a ……………(i) 
Tq = b 
⇒ A + (q – 1) d = b …………….(ii) 
and Tr = c 
⇒ A + (r – 1) d = c ……………..(iii) 
(i) On multiplying eq. (i) by (q – r), eq. (ii) by (r – p) and eq. (iii) by (p – q),we get 
(q – r) A + (p – 1) (q – r) d = a (q – r) …………..(iv) 
(r – p) A + (q – 1) (r – p) d = b (r – p) …………….(v) 
and (p – q) A + (r – 1) (p – q) d = c (p – q) ……………(vi) 
On adding eqs. (iv), (v) and eq. (vi), we get 
(q – r) A + (p – 1) (q – r) d + (r – p) A + (q – 1) (r – p) d + (p – q) A + (r – 1) (p – q) d = a (q – r) + b (r – p) + c (p – q) 
⇒ A [(q – r) + (r – p) + (p – q)] + (p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)] d = a (q – r) + b (r – p) + c (p – q) 
A(0) + (0)d = a (q – r) + b (r – p) + c (p – q) 
a (q – r) + b (r – p) + c (p – q) = 0 
Hence proved.
Miscellaneous Exercise Class 11 Maths Question 16: 
If a(1b+1c),b(1c+1a),c(1a+1b)<script type=”math/tex” id=”MathJax-Element-168″>a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)</script> are in A.P.,prove that a, b, c are in A.P. 
Ans:
It is given that a(1b+1c),b(1c+1a),c(1a+1b)<script type=”math/tex” id=”MathJax-Element-169″>a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)</script> are in A.P.
∴ b(1c+1a)−a(1b+1c)=c(1a+1b)−b(1c+1a)<script type=”math/tex” id=”MathJax-Element-170″>b\left(\frac{1}{c}+\frac{1}{a}\right)-a\left(\frac{1}{b}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)-b\left(\frac{1}{c}+\frac{1}{a}\right)</script>
⇒ b(a+c)ac−a(b+c)bc=c(a+b)ab−b(a+c)ac<script type=”math/tex” id=”MathJax-Element-171″>\frac{b(a+c)}{a c}-\frac{a(b+c)}{b c}=\frac{c(a+b)}{a b}-\frac{b(a+c)}{a c}</script>
⇒ b2a+b2c−a2b−a2cabc=c2a+c2b−b2a−b2cabc<script type=”math/tex” id=”MathJax-Element-172″>\frac{b^{2} a+b^{2} c-a^{2} b-a^{2} c}{a b c}=\frac{c^{2} a+c^{2} b-b^{2} a-b^{2} c}{a b c}</script>
⇒ b2a – a2b + b2c – a2c = c2a – b2a + c2b – b2c 
⇒ ab (b – a) + c (b2 – a2) = a (c2 – b2) + bc (c – b) 
⇒ ab (b – a) + c (b – a) (b + a) = a (c – b) (c + b) + bc (c – b) 
⇒ (b – a) (ab + cb + ca) = (c – b) (ac + ab + bc) 
⇒ b – a = c – b 
Thus, a, b and c are in A.P.
Miscellaneous Exercise Class 11 Maths Question 17: 
If a, b, c, d are in G.P. prove that (an + bn), (bn + cn), (cn + dn) are in G.P. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120487″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q17.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q17.1″ width=”524″ height=”471″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q17.1.png 524w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q17.1-300×270.png 300w” sizes=”(max-width: 524px) 100vw, 524px”>
Miscellaneous Exercise Class 11 Maths Question 18: 
If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17 : 15. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120489″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q18.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q18.1″ width=”415″ height=”661″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q18.1.png 415w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q18.1-188×300.png 188w” sizes=”(max-width: 415px) 100vw, 415px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120490″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q18.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q18.2″ width=”450″ height=”509″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q18.2.png 450w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q18.2-265×300.png 265w” sizes=”(max-width: 450px) 100vw, 450px”>
Miscellaneous Exercise Class 11 Maths Question 19: 
The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that a : b (m + m2−n2−−−−−−−√<script type=”math/tex” id=”MathJax-Element-173″>\sqrt{m^{2}-n^{2}}</script>) : (m – m2−n2−−−−−−−√<script type=”math/tex” id=”MathJax-Element-174″>\sqrt{m^{2}-n^{2}}</script>). 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120492″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q19.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q19.1″ width=”464″ height=”636″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q19.1.png 464w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q19.1-219×300.png 219w” sizes=”(max-width: 464px) 100vw, 464px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120493″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q19.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q19.2″ width=”349″ height=”328″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q19.2.png 349w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q19.2-300×282.png 300w” sizes=”(max-width: 349px) 100vw, 349px”>
Miscellaneous Exercise Class 11 Maths Question 20: 
<img loading=”lazy” class=”alignnone size-full wp-image-120497″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q20.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q20″ width=”557″ height=”54″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q20.png 557w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q20-300×29.png 300w” sizes=”(max-width: 557px) 100vw, 557px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120495″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q20.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q20.1″ width=”424″ height=”554″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q20.1.png 424w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q20.1-230×300.png 230w” sizes=”(max-width: 424px) 100vw, 424px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120496″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q20.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q20.2″ width=”328″ height=”336″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q20.2.png 328w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q20.2-293×300.png 293w” sizes=”(max-width: 328px) 100vw, 328px”>
Miscellaneous Exercise Class 11 Maths Question 21: 
Find the sum of the following series up to n terms. 
(i) 5 + 55 + 555 + …………… 
(ii) .6 + .66 + .666 + …………… 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120498″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q21.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q21.1″ width=”357″ height=”647″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q21.1.png 357w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q21.1-166×300.png 166w” sizes=”(max-width: 357px) 100vw, 357px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120499″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q21.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q21.2″ width=”171″ height=”196″>
Miscellaneous Exercise Class 11 Maths Question 22: 
<img loading=”lazy” class=”alignnone size-full wp-image-120502″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q22.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q22″ width=”445″ height=”28″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q22.png 445w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q22-300×19.png 300w” sizes=”(max-width: 445px) 100vw, 445px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120501″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q22.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q22.1″ width=”407″ height=”143″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q22.1.png 407w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q22.1-300×105.png 300w” sizes=”(max-width: 407px) 100vw, 407px”>
Miscellaneous Exercise Class 11 Maths Question 23: 
<img loading=”lazy” class=”alignnone size-full wp-image-120505″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q23.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q23″ width=”472″ height=”28″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q23.png 472w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q23-300×18.png 300w” sizes=”(max-width: 472px) 100vw, 472px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120503″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q23.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q23.1″ width=”529″ height=”546″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q23.1.png 529w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q23.1-291×300.png 291w” sizes=”(max-width: 529px) 100vw, 529px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120504″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q23.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q23.2″ width=”279″ height=”293″>
Miscellaneous Exercise Class 11 Maths Question 24: 
<img loading=”lazy” class=”alignnone size-full wp-image-120507″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q24.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q24″ width=”521″ height=”50″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q24.png 521w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q24-300×29.png 300w” sizes=”(max-width: 521px) 100vw, 521px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120506″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q24.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q24.1″ width=”429″ height=”541″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q24.1.png 429w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q24.1-238×300.png 238w” sizes=”(max-width: 429px) 100vw, 429px”>
Miscellaneous Exercise Class 11 Maths Question 25: 
<img loading=”lazy” class=”alignnone size-full wp-image-120509″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q25.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q25″ width=”517″ height=”50″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q25.png 517w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q25-300×29.png 300w” sizes=”(max-width: 517px) 100vw, 517px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120508″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q25.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q25.1″ width=”622″ height=”466″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q25.1.png 622w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q25.1-300×225.png 300w” sizes=”(max-width: 622px) 100vw, 622px”>
Miscellaneous Exercise Class 11 Maths Question 26: 
<img loading=”lazy” class=”alignnone size-full wp-image-120512″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q26.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q26″ width=”330″ height=”58″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q26.png 330w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q26-300×53.png 300w” sizes=”(max-width: 330px) 100vw, 330px”> 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120510″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q26.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q26.1″ width=”483″ height=”562″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q26.1.png 483w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q26.1-258×300.png 258w” sizes=”(max-width: 483px) 100vw, 483px”>
<img loading=”lazy” class=”alignnone size-full wp-image-120511″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q26.2.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q26.2″ width=”478″ height=”590″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q26.2.png 478w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q26.2-243×300.png 243w” sizes=”(max-width: 478px) 100vw, 478px”>
Miscellaneous Exercise Class 11 Maths Question 27: 
A farmer buys a used tractor for Rs. 12000. He pays Rs. 6000 cash and agrees to pay the balance in annual installments of Rs. 500 plus 12% interest on the unpaid amount. How much will be the tractor cost him? 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120513″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q27.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q27.1″ width=”578″ height=”660″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q27.1.png 578w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q27.1-263×300.png 263w” sizes=”(max-width: 578px) 100vw, 578px”>
Miscellaneous Exercise Class 11 Maths Question 28: 
Shamshad Ali buys a scooter for Rs. 22000. He pays Rs. 4000 cash and agrees to pay the balance in annual installment of Rs. 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him? 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120515″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q28.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q28.1″ width=”591″ height=”682″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q28.1.png 591w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q28.1-260×300.png 260w” sizes=”(max-width: 591px) 100vw, 591px”>
Miscellaneous Exercise Class 11 Maths Question 29: 
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120517″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q29.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q29.1″ width=”515″ height=”439″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q29.1.png 515w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q29.1-300×256.png 300w” sizes=”(max-width: 515px) 100vw, 515px”>
Miscellaneous Exercise Class 11 Maths Question 30: 
A man deposited Rs. 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120519″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q30.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q30.1″ width=”548″ height=”458″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q30.1.png 548w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q30.1-300×251.png 300w” sizes=”(max-width: 548px) 100vw, 548px”>
Miscellaneous Exercise Class 11 Maths Question 31: 
A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120521″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q31.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q31.1″ width=”577″ height=”269″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q31.1.png 577w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q31.1-300×140.png 300w” sizes=”(max-width: 577px) 100vw, 577px”>
Miscellaneous Exercise Class 11 Maths Question 32: 
150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed. 
Ans: 
<img loading=”lazy” class=”alignnone size-full wp-image-120523″ src=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q32.1.png” alt=”NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Ex Q32.1″ width=”556″ height=”684″ srcset=”https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q32.1.png 556w, https://www.learncbse.in/wp-content/uploads/2020/11/NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Miscellaneous-Ex-Q32.1-244×300.png 244w” sizes=”(max-width: 556px) 100vw, 556px”>
<div class=”featured-snippet”>
<h3>NCERT Solutions for Class 11 Maths All Chapters</h3>
<ul>
<li><a title=”CBSE Maths Solutions for Class 11th Chapter 1 Sets” href=”https://www.learncbse.in/cbse-maths-solutions-class-11th-chapter-1-sets/” target=”_blank” rel=”noopener noreferrer”>Chapter 1 Sets</a></li>
<li><a title=”NCERT Class 11 Solutions Maths Chapter 2 Relations and Functions” href=”https://www.learncbse.in/ncert-class-11-solutions-maths-chapter-2-relations-functions/” target=”_blank” rel=”noopener noreferrer”>Chapter 2 Relations and Functions</a></li>
<li><a title=”CBSE FREE NCERT Solutions for Class 11th Maths Chapter 3 Trigonometric Functions” href=”https://www.learncbse.in/cbse-free-ncert-solutions-for-class-11th-maths-chapter-3-trigonometric-functions/” target=”_blank” rel=”noopener noreferrer”>Chapter 3 Trigonometric Functions</a></li>
<li><a title=”NCERT Solutions for Class 11th Maths Chapter 4 Principle of Mathematical Induction” href=”https://www.learncbse.in/ncert-solutions-class-11th-maths-chapter-4-principle-mathematical-induction/” target=”_blank” rel=”noopener noreferrer”>Chapter 4 Principle of Mathematical Induction</a></li>
<li><a title=”NCERT Solutions for Class 11th Maths Chapter 5 Complex Numbers and Quadratic Equations” href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-5-complex-numbers-and-quadratic-equations/” target=”_blank” rel=”noopener noreferrer”>Chapter 5 Complex Numbers and Quadratic Equations</a></li>
<li><a title=”FREE NCERT Solutions for Class 11th Maths Chapter 6 Linear Inequalities” href=”https://www.learncbse.in/free-ncert-solutions-for-class-11th-maths-chapter-6-linear-inequalities/” target=”_blank” rel=”noopener noreferrer”>Chapter 6 Linear Inequalities</a></li>
<li><a title=”NCERT Solutions for Class 11th Maths Chapter 7 Permutation and Combinations” href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-7-permutation-and-combinations/” target=”_blank” rel=”noopener noreferrer”>Chapter 7 Permutation and Combinations</a></li>
<li><a title=”NCERT Solutions for Class 11th Maths Chapter 8 Binomial Theorem” href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-8-binomial-theorem/” target=”_blank” rel=”noopener noreferrer”>Chapter 8 Binomial Theorem</a></li>
<li><a title=”NCERT Solutions for Class 11th Maths Chapter 9 Sequences and Series” href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-9-sequences-and-series/” target=”_blank” rel=”noopener noreferrer”>Chapter 9 Sequences and Series</a></li>
<li><a title=”FREE NCERT Solutions for Class 11th Maths Chapter 10 Straight Lines” href=”https://www.learncbse.in/free-ncert-solutions-class-11th-maths-chapter-10-straight-lines/” target=”_blank” rel=”noopener noreferrer”>Chapter 10 Straight Lines</a></li>
<li><a title=”NCERT Solutions for Class 11th Maths Chapter 11 Conic Sections” href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-11-conic-sections/” target=”_blank” rel=”noopener noreferrer”>Chapter 11 Conic Sections</a></li>
<li><a title=”NCERT Solutions for Class 11th Maths Chapter 12 Introduction to three Dimensional Geometry” href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-12-introduction-to-three-dimensional-geometry/” target=”_blank” rel=”noopener noreferrer”>Chapter 12 Introduction to Three Dimensional Geometry</a></li>
<li><a title=”NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives” href=”https://www.learncbse.in/ncert-solutions-class-11-maths-chapter-13-limits-derivatives/” target=”_blank” rel=”noopener noreferrer”>Chapter 13 Limits and Derivatives</a></li>
<li><a title=”NCERT Solutions for Class 11th Maths Chapter 14 Mathematical Reasoning” href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-14-mathematical-reasoning/” target=”_blank” rel=”noopener noreferrer”>Chapter 14 Mathematical Reasoning</a></li>
<li><a title=”NCERT Solutions for Class 11th Maths Chapter 15 Statistics” href=”https://www.learncbse.in/ncert-solutions-class-11th-maths-chapter-15-statistics/” target=”_blank” rel=”noopener noreferrer”>Chapter 15 Statistics</a></li>
<li><a title=”NCERT Solutions for Class 11th Maths Chapter 16 Probability” href=”https://www.learncbse.in/ncert-solutions-for-class-11th-maths-chapter-16-probability/” target=”_blank” rel=”noopener noreferrer”>Chapter 16 Probability</a></li>
</ul>
</div>
</div>

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

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NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

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