Ch. - 9 Sequences and Series

Ch. - 9 Sequences and Series

Chapter 9 Ex.9.1 Question 1

Write the first five terms of the sequences whose \({n^{th}}\)term is \({a_n} = n\left( {n + 2} \right)\).

Solution

\[{a_n} = n\left( {n + 2} \right)\]

Substituting \(n = 1,2,3,4,5\)

\[\begin{align}{a_1}&= 1\left( {1 + 2} \right) = 3\\{a_2} &= 2\left( {2 + 2} \right) = 8\\{a_3} &= 3\left( {3 + 2} \right) = 15\\{a_4}& = 4\left( {4 + 2} \right) = 24\\{a_5} &= 5\left( {5 + 2} \right) = 35\end{align}\]

Therefore, the required terms are \(3,8,15,24\) and \(35\).

Chapter 9 Ex.9.1 Question 2

Write the first five terms of the sequences whose \({n^{th}}\) term is \({a_n} = \frac{n}{{n + 1}}\).

Solution

\[{a_n} = \frac{n}{{n + 1}}\]

Substituting \(n = 1,2,3,4,5\)

\[\begin{align}{a_1} &= \frac{1}{{1 + 1}} = \frac{1}{2}\\{a_2} &= \frac{2}{{2 + 1}} = \frac{2}{3}\\{a_3} &= \frac{3}{{3 + 1}} = \frac{3}{4}\\{a_4}& = \frac{4}{{4 + 1}} = \frac{4}{5}\\{a_5}& = \frac{5}{{5 + 1}} = \frac{5}{6}\end{align}\]

Therefore, the required terms are \(\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}\) and \(\frac{5}{6}\).

Chapter 9 Ex.9.1 Question 3

Write the first five terms of the sequences whose \({n^{th}}\) term is \({a_n} = {2^n}\).

Solution

\[{a_n} = {2^n}\]

Substituting \(n = 1,2,3,4,5\)

\[\begin{align}{a_1} &= {2^1} = 2\\{a_2}& = {2^2} = 4\\{a_3} &= {2^3} = 8\\{a_4} &= {2^4} = 16\\{a_5} &= {2^5} = 32\end{align}\]

Therefore, the required terms are \(2,4,8,16\) and \(32\).

Chapter 9 Ex.9.1 Question 4

Write the first five terms of the sequences whose \({n^{th}}\) term is \({a_n} = \frac{{2n - 3}}{6}\).

Solution

\[{a_n} = \frac{{2n - 3}}{6}\]

Substituting \(n = 1,2,3,4,5\)

\[\begin{align}{a_1} &= \frac{{2\left( 1 \right) - 3}}{6} = \frac{{ - 1}}{6}\\{a_2} &= \frac{{2\left( 2 \right) - 3}}{6} = \frac{1}{6}\\{a_3} &= \frac{{2\left( 3 \right) - 3}}{6} = \frac{1}{2}\\{a_4} &= \frac{{2\left( 4 \right) - 3}}{6} = \frac{5}{6}\\{a_5} &= \frac{{2\left( 5 \right) - 3}}{6} = \frac{7}{6}\end{align}\]

Therefore, the required terms are \(\frac{{ - 1}}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6}\) and \(\frac{7}{6}\).

Chapter 9 Ex.9.1 Question 5

Write the first five terms of the sequences whose \({n^{th}}\)term is \({a_n} = {\left( { - 1} \right)^{n - 1}}{5^{n + 1}}\).

Solution

\[{a_n} = {\left( { - 1} \right)^{n - 1}}{5^{n + 1}}\]

Substituting \(n = 1,2,3,4,5\)

\[\begin{align}{a_1} &= {\left( { - 1} \right)^{1 - 1}}{5^{1 + 1}} = {5^2} = 25\\{a_2} &= {\left( { - 1} \right)^{2 - 1}}{5^{2 + 1}} = - {5^3} = - 125\\{a_3} &= {\left( { - 1} \right)^{3 - 1}}{5^{3 + 1}} = {5^4} = 625\\{a_4} &= {\left( { - 1} \right)^{4 - 1}}{5^{4 + 1}} = - {5^5} = - 3125\\{a_5} &= {\left( { - 1} \right)^{5 - 1}}{5^{5 + 1}} = {5^6} = 15625\end{align}\]

Therefore, the required terms are \(25, - 125,625, - 3125\) and \(15625\).

Chapter 9 Ex.9.1 Question 6

Write the first five terms of the sequences whose \({n^{th}}\) term is \({a_n} = n\frac{{{n^2} + 5}}{4}\).

Solution

\[{a_n} = n\frac{{{n^2} + 5}}{4}\]

Substituting \(n = 1,2,3,4,5\)

\[\begin{align}{a_1}& = 1\frac{{{1^2} + 5}}{4} = \frac{3}{2}\\{a_2} &= 2\frac{{{2^2} + 5}}{4} = \frac{9}{2}\\{a_3} &= 3\frac{{{3^2} + 5}}{4} = \frac{{21}}{2}\\{a_4} &= 4\frac{{{4^2} + 5}}{4} = 21\\{a_5} &= 5\frac{{{5^2} + 5}}{4} = \frac{{75}}{2}\end{align}\]

Therefore, the required terms are \(\frac{3}{2},\frac{9}{2},\frac{{21}}{2},21\) and \(\frac{{75}}{2}\).

Chapter 9 Ex.9.1 Question 7

Find the \({17^{th}}\) and \({24^{th}}\) term of the sequences whose \({n^{th}}\)term is \({a_n} = 4n - 3\).

Solution

\[{a_n} = 4n - 3\]

Substituting \(n=17\)

\[{a_{17}} = 4\left( {17} \right) - 3 = 68 - 3 = 65\]

Substituting \(n=24\)

\[{{a}_{24}}=4\left( 24 \right)-3=96-3=93\]

Chapter 9 Ex.9.1 Question 8

Write the \({7^{th}}\) term of the sequences whose \({n^{th}}\)term is \({a_n} = \frac{{{n^2}}}{{{2^n}}}\).

Solution

\[{a_n} = \frac{{{n^2}}}{{{2^n}}}\]

Substituting \(n = 7\)

\[{a_7} = \frac{{{7^2}}}{{{2^7}}} = \frac{{49}}{{128}}\]

Chapter 9 Ex.9.1 Question 9

Write the \({9^{th}}\) term of the sequences whose \({n^{th}}\)term is \({a_n} = {\left( { - 1} \right)^{n - 1}}{n^3}\).

Solution

\[{a_n} = {\left( { - 1} \right)^{n - 1}}{n^3}\]

Substituting \(n = 9\)

\[{a_9} = {\left( { - 1} \right)^{9 - 1}}{9^3} = 729\]

Chapter 9 Ex.9.1 Question 10

Write the \({20^{th}}\) term of the sequences whose \({n^{th}}\)term is \({a_n} = \frac{{n\left( {n - 2} \right)}}{{n + 3}}\).

Solution

\[{a_n} = \frac{{n\left( {n - 2} \right)}}{{n + 3}}\]

Substituting \(n = 20\)

\[{a_{20}} = \frac{{20\left( {20 - 2} \right)}}{{20 + 3}} = \frac{{360}}{{23}}\]

Chapter 9 Ex.9.1 Question 11

Write the first five terms of the following sequence and obtain the corresponding series:\({a_1} = 3,{a_n} = 3{a_{n - 1}} + 2\)for all \(n > 1\).

Solution

\({a_1} = 3,{a_n} = 3{a_{n - 1}} + 2\) for all \(n > 1\).

\[\begin{align}{a_2}& = 3{a_1} + 2 = 3\left( 3 \right) + 2 = 11\\{a_3} &= 3{a_2} + 2 = 3\left( {11} \right) + 2 = 35\\{a_4} &= 3{a_3} + 2 = 3\left( {35} \right) + 2 = 107\\{a_5} &= 3{a_4} + 2 = 3\left( {107} \right) + 2 = 323\end{align}\]

Hence, the first five terms of the sequence are \(3,11,35,107\) and \(323\).

The corresponding series is \(3 + 11 + 35 + 107 + 323 + \ldots \)

Chapter 9 Ex.9.1 Question 12

Write the first five terms of the following sequence and obtain the corresponding series:\({a_1} = - 1,{a_n} = \frac{{{a_{n - 1}}}}{n}\) for all \(n \ge 2\).

Solution

\[{a_1} = - 1,\;{a_n} = \frac{{{a_{n - 1}}}}{n};\;\;n \ge 2\]

\[\begin{align}{a_2}&= \frac{{{a_1}}}{2} = \frac{{ - 1}}{2}\\{a_3}&= \frac{{{a_2}}}{3} = \frac{{ - 1}}{6}\\{a_4} &= \frac{{{a_3}}}{4} = \frac{{ - 1}}{{24}}\\{a_5}&= \frac{{{a_4}}}{5} = \frac{{ - 1}}{{120}}\end{align}\]

Hence, the first five terms of the sequence are \( - 1,\frac{{ - 1}}{2},\frac{{ - 1}}{6},\frac{{ - 1}}{{24}}\) and \(\frac{{ - 1}}{{120}}\).

The corresponding series is \(\left( { - 1} \right) + \left( {\frac{{ - 1}}{2}} \right) + \left( {\frac{{ - 1}}{6}} \right) + \left( {\frac{{ - 1}}{{24}}} \right) + \left( {\frac{{ - 1}}{{120}}} \right) + \ldots \)

Chapter 9 Ex.9.1 Question 13

Write the first five terms of the following sequence and obtain the corresponding series:\({a_1} = {a_2} = 2,{a_n} = {a_{n - 1}} - 1,n > 2\).

Solution

\[{a_1} = {a_2} = 2,{a_n} = {a_{n - 1}} - 1,n > 2\]

\[\begin{align}\quad\Rightarrow {a_3}&= {a_2} - 1 = 2 - 1 = 1\\{a_4}&= {a_3} - 1 = 1 - 1 = 0\\{a_5}&= {a_4} - 1 = 0 - 1 = - 1\end{align}\]

Hence, the first five terms of the sequence are \(2,2,1,0\) and \( - 1\)

The corresponding series is \(2 + 2 + 1 + 0 + \left( { - 1} \right) + \ldots \)

Chapter 9 Ex.9.1 Question 14

The Fibonacci sequence is defined by \(1 = {a_1} = {a_2}\)and \({a_n} = {a_{n - 1}} + {a_{n - 2}},n > 2\). Find \(\frac{{{a_{n + 1}}}}{{{a_n}}}\), for \(n = 1,2,3,4,5\)

Solution

\[1 = {a_1} = {a_2}\]

\[{a_n} = {a_{n - 1}} + {a_{n - 2}},n > 2\]

Therefore,

\[\begin{align}{a_3} &= {a_2} + {a_1} = 1 + 1 = 2\\{a_4} &= {a_3} + {a_2} = 2 + 1 = 3\\{a_5} &= {a_4} + {a_3} = 3 + 2 = 5\\{a_6} &= {a_5} + {a_4} = 5 + 3 = 8\end{align}\]

For \(n = 1,\;\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_2}}}{{{a_1}}} = \frac{1}{1} = 1\)

For \(n = 2,\;\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_3}}}{{{a_2}}} = \frac{2}{1} = 2\)

For \(n = 3,\;\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_4}}}{{{a_3}}} = \frac{3}{2}\)

For \(n = 4,\;\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_5}}}{{{a_4}}} = \frac{5}{3}\)

For \(n = 5,\;\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_6}}}{{{a_5}}} = \frac{8}{5}\)

Download FREE PDF of Chapter-9 Sequences and Series
Sequences and Series | NCERT Solutions
Download Cuemath NCERT App

Learn from the best math teachers and top your exams

Learn from the best

math teachers and top

your exams


Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program
  
Download Cuemath NCERT App
Related Sections
Related Sections

Learn from the best math teachers and top your exams

Learn from the best

math teachers and top

your exams


Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program