# NCERT Solutions For Class 11 Maths Chapter 9 Exercise 9.1

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## 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 $$n = 5,\;\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_6}}}{{{a_5}}} = \frac{8}{5}$$

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