NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series - Exercise 9.1

Exercise 9.1

1. Write the first five terms of the sequences whose ${{\text{n}}^{\text{th}}}$ term is \[{{\text{a}}_{n}}\text{= n(n+2)}\]. 

Ans:

First we will use the given sequence to find the required terms by putting $\text{n = 1,2,3,4,5}\text{.}$ 

Here we have 

For $\text{n = 1}$

     \[{{\text{a}}_{1}}\text{= 1(1+2)}\] 

\[\Rightarrow {{\text{a}}_{\text{1}}}\text{= 3}\]

For $\text{n = 2}$

     \[{{\text{a}}_{\text{2}}}\text{= 2(2+2)}\]

\[\Rightarrow {{\text{a}}_{\text{2}}}\text{= 8}\]

For $\text{n = 3}$

     \[{{\text{a}}_{\text{3}}}\text{= 3(3+2)}\]

\[\Rightarrow {{\text{a}}_{\text{3}}}\text{= 15}\]

For  $\text{n = 4}$

     \[{{\text{a}}_{\text{4}}}\text{= 4(4+2)}\]

\[\Rightarrow {{\text{a}}_{\text{4}}}\text{= 24}\]

For $\text{n = 5}$

     \[{{\text{a}}_{\text{5}}}\text{= 5(5+2)}\]

\[\Rightarrow {{\text{a}}_{\text{5}}}\text{= 35}\]

Hence the required first five terms of given sequence are $\text{2,8,15,24,35}$ .

2. Write the first five terms of the sequences whose ${{\text{n}}^{\text{th}}}$ term is \[{{\text{a}}_{\text{n}}}\text{=}\frac{\text{n}}{\text{n+1}}\]. 

Ans:

First we will use the given sequence to find the required terms by putting $\text{n = 1,2,3,4,5}\text{.}$ 

Here we have 

For $\text{n = 1}$

\[{{\text{a}}_{\text{1}}}\text{= }\frac{\text{1}}{\text{1+1}}\] 

\[{{\text{a}}_{\text{1}}}\text{= 1}\]

For $\text{n = 2}$

\[{{\text{a}}_{\text{2}}}\text{= }\frac{\text{2}}{\text{2+1}}\]

\[{{\text{a}}_{\text{2}}}\text{= }\frac{\text{2}}{\text{3}}\]

For $\text{n = 3}$

\[{{\text{a}}_{\text{3}}}\text{= }\frac{\text{3}}{\text{3+1}}\]

\[{{\text{a}}_{\text{3}}}\text{= }\frac{\text{3}}{\text{4}}\]

For $\text{n = 4}$

\[{{\text{a}}_{\text{4}}}\text{= }\frac{\text{4}}{\text{4+1}}\]

\[{{\text{a}}_{\text{4}}}\text{= }\frac{\text{4}}{\text{5}}\]

For $\text{n = 5}$

\[{{\text{a}}_{\text{5}}}\text{= }\frac{\text{5}}{\text{5+1}}\]

\[{{\text{a}}_{\text{5}}}\text{= }\frac{\text{5}}{\text{6}}\]

Hence the required first five terms of given sequence are  \[\frac{\text{1}}{\text{2}}\text{,}\frac{\text{2}}{\text{3}}\text{,}\frac{\text{3}}{\text{4}}\text{,}\frac{\text{4}}{\text{5}}\text{,}\frac{\text{5}}{\text{6}}\text{.}\]

3. Write the first five terms of the sequences whose ${{\text{n}}^{\text{th}}}$ term is \[{{\text{a}}_{n}}\text{= }{{\text{2}}^{\text{n}}}\]. 

Ans:

First we will use the given sequence to find the required terms by putting $\text{n = 1,2,3,4,5}\text{.}$ 

Here we have 

For $\text{n = 1}$

\[{{\text{a}}_{\text{1}}}\text{= }{{\text{2}}^{\text{1}}}\] 

\[{{\text{a}}_{\text{1}}}\text{= 2}\]

For $\text{n = 2}$

\[{{\text{a}}_{\text{2}}}\text{= }{{\text{2}}^{\text{2}}}\]

\[{{\text{a}}_{\text{2}}}\text{= 4}\]

For $\text{n = 3}$

\[{{\text{a}}_{\text{3}}}\text{= }{{\text{2}}^{\text{3}}}\]

\[{{\text{a}}_{\text{3}}}\text{= 8}\]

For $\text{n = 4}$

\[{{\text{a}}_{\text{4}}}\text{= }{{\text{2}}^{\text{4}}}\]

\[{{\text{a}}_{\text{4}}}\text{= 16}\]

For  $\text{n = 5}$

\[{{\text{a}}_{\text{5}}}\text{= }{{\text{2}}^{\text{5}}}\]

\[{{\text{a}}_{\text{5}}}\text{= 32}\]

Hence the required first five terms of given sequence are $\text{2,4,8,16,32}\text{.}$ 

4. Write the first five terms of the sequences whose ${{\text{n}}^{\text{th}}}$  term is \[{{\text{a}}_{\text{n}}}\text{= }\frac{\text{2n-3}}{\text{6}}\]. 

Ans: First we will use the given sequence to find the required terms by putting $\text{n = 1,2,3,4,5}\text{.}$ 

Here we have 

For $\text{n = 1}$

\[{{\text{a}}_{\text{1}}}\text{= }\frac{\text{2(1) - 3}}{\text{6}}\] 

\[{{\text{a}}_{\text{1}}}\text{= }\frac{\text{-1}}{\text{6}}\]

For $\text{n = 2}$

\[{{\text{a}}_{\text{2}}}\text{= }\frac{\text{2(2) - 3}}{\text{6}}\]

\[{{\text{a}}_{\text{2}}}\text{= }\frac{\text{1}}{\text{6}}\]

For $\text{n = 3}$

     \[{{\text{a}}_{\text{3}}}\text{= }\frac{\text{2(3) - 3}}{\text{6}}\]

     \[{{\text{a}}_{\text{3}}}\text{= }\frac{\text{3}}{\text{6}}\]

\[\Rightarrow {{\text{a}}_{\text{3}}}\text{= }\frac{\text{1}}{\text{2}}\]

For $\text{n = 4}$

\[{{\text{a}}_{\text{4}}}\text{= }\frac{\text{2(4) - 3}}{\text{6}}\]

\[{{\text{a}}_{\text{4}}}\text{= }\frac{\text{5}}{\text{6}}\]

For $\text{n = 5}$

\[{{\text{a}}_{\text{5}}}\text{= }\frac{\text{2(5) - 3}}{\text{6}}\]

\[{{\text{a}}_{\text{5}}}\text{= }\frac{\text{7}}{\text{6}}\]

Hence the required first five terms of given sequence are  \[\text{-}\frac{\text{1}}{\text{6}}\text{,}\frac{\text{1}}{\text{6}}\text{,}\frac{\text{1}}{\text{2}}\text{,}\frac{\text{5}}{\text{6}}\text{,}\frac{\text{7}}{\text{6}}\text{.}\]

5. Write the first five terms of the sequences whose term is \[{{\text{a}}_{\text{n}}}\text{= (-1}{{\text{)}}^{\text{n-1}}}{{\text{5}}^{\text{n+1}}}\]. 

Ans:

First we will use the given sequence to find the required terms by putting $\text{n = 1,2,3,4,5}\text{.}$ 

Here we have 

For  $\text{n = 1}$

     \[{{\text{a}}_{\text{1}}}\text{=  (-1}{{\text{)}}^{\text{(1-1)}}}{{\text{5}}^{\text{(1+1)}}}\] 

        \[\text{= (-1}{{\text{)}}^{\text{0}}}{{\text{(5)}}^{\text{2}}}\]

\[\Rightarrow {{\text{a}}_{\text{1}}}\text{=  25}\]

For  $\text{n = 2}$

     \[{{\text{a}}_{\text{2}}}\text{= (-1}{{\text{)}}^{\text{(2-1)}}}{{\text{5}}^{\text{(2+1)}}}\]

          \[\text{= (-1}{{\text{)}}^{\text{1}}}{{\text{(5)}}^{\text{3}}}\]

\[\Rightarrow {{\text{a}}_{\text{2}}}\text{= -125}\]

For $\text{n = 3}$

     \[{{\text{a}}_{\text{3}}}\text{= (-1}{{\text{)}}^{\text{(3-1)}}}{{\text{5}}^{\text{(3+1)}}}\]

         \[\text{= (-1}{{\text{)}}^{\text{2}}}{{\text{(5)}}^{\text{4}}}\]

\[\Rightarrow {{\text{a}}_{\text{3}}}\text{= 625}\]

For $\text{n = 4}$

      \[{{\text{a}}_{\text{4}}}\text{=  (-1}{{\text{)}}^{\text{(4-1)}}}{{\text{5}}^{\text{(4+1)}}}\]

         \[\text{= (-1}{{\text{)}}^{\text{3}}}{{\text{(5)}}^{\text{5}}}\]

\[\Rightarrow {{\text{a}}_{\text{4}}}\text{= -3125}\]

For $\text{n = 5}$

     \[{{\text{a}}_{\text{5}}}\text{= (-1}{{\text{)}}^{\text{(5-1)}}}{{\text{5}}^{\text{(5+1)}}}\]

         \[\text{= (-1}{{\text{)}}^{\text{4}}}{{\text{(5)}}^{\text{6}}}\]

\[\Rightarrow {{\text{a}}_{\text{5}}}\text{= 15625}\]

Hence the required first five terms of given sequence are  $\text{25,-125,625,-3125,15625}\text{.}$

6. Write the first five terms of the sequences whose ${{\text{n}}^{\text{th}}}$ term is \[{{\text{a}}_{\text{n}}}\text{= n}\frac{{{\text{n}}^{\text{2}}}\text{+5}}{\text{4}}\]. 

Ans:

First we will use the given sequence to find the required terms by putting $\text{n = 1,2,3,4,5}\text{.}$ 

Here we have 

For $\text{n = 1}$

     \[{{\text{a}}_{\text{1}}}\text{= 1 }\!\!\times\!\!\text{ }\frac{{{\text{1}}^{\text{2}}}\text{+5}}{\text{4}}\] 

        \[\text{= }\frac{\text{6}}{\text{4}}\]

\[\Rightarrow {{\text{a}}_{\text{1}}}\text{= }\frac{\text{3}}{\text{2}}\]

For  $\text{n = 2}$

     \[{{\text{a}}_{\text{2}}}\text{= 2 }\!\!\times\!\!\text{ }\frac{{{\text{2}}^{\text{2}}}\text{+5}}{\text{4}}\]

          \[\text{= 2 }\!\!\times\!\!\text{ }\frac{\text{9}}{\text{4}}\]

\[\Rightarrow {{\text{a}}_{\text{2}}}\text{= }\frac{\text{9}}{\text{2}}\]

For $\text{n = 3}$

     \[{{\text{a}}_{\text{3}}}\text{= 3 }\!\!\times\!\!\text{ }\frac{{{\text{3}}^{\text{2}}}\text{+5}}{\text{4}}\]

         \[\text{= 3 }\!\!\times\!\!\text{ }\frac{14}{\text{4}}\]

\[\Rightarrow {{\text{a}}_{\text{3}}}\text{= }\frac{\text{21}}{\text{2}}\]

For  $\text{n = 4}$

      \[{{\text{a}}_{\text{4}}}\text{=  4 }\!\!\times\!\!\text{ }\frac{{{\text{4}}^{\text{2}}}\text{+5}}{\text{4}}\]

\[\Rightarrow {{\text{a}}_{\text{4}}}\text{= 21}\]

For $\text{n = 5}$

      \[{{\text{a}}_{\text{5}}}\text{= 5 }\!\!\times\!\!\text{ }\frac{{{\text{5}}^{\text{2}}}\text{+5}}{\text{4}}\]

         \[\text{= 5 }\!\!\times\!\!\text{ }\frac{\text{30}}{\text{4}}\]

\[\Rightarrow {{\text{a}}_{\text{5}}}\text{= }\frac{\text{75}}{\text{2}}\]

Hence the required first five terms of given sequence are  $\frac{\text{3}}{\text{2}}\text{,}\frac{\text{9}}{\text{2}}\text{,}\frac{\text{21}}{\text{2}}\text{,21,}\frac{\text{75}}{\text{2}}\text{.}$

7. Find the $\text{1}{{\text{7}}^{\text{th}}}$ and $\text{1}{{\text{8}}^{\text{th}}}$ term in the following sequence whose ${{\text{n}}^{\text{th}}}$ term is ${{\text{a}}_{\text{n}}}\text{=4n-3}$

Ans:

To find $\text{1}{{\text{7}}^{\text{th}}}$  term we will substitute $\text{n}$ by $\text{n =17}$

Here we have 

      ${{\text{a}}_{\text{17}}}\text{= 4(17)-3}$

      ${{\text{a}}_{\text{17}}}\text{=65-3}$

$\Rightarrow {{\text{a}}_{\text{17}}}\text{= 65}$

To find ${{18}^{th}}$  term we will substitute $\text{n}$ by $\text{n =18}$

Here we have 

      ${{\text{a}}_{\text{24}}}\text{= 4(24)-3}$

      ${{\text{a}}_{\text{24}}}\text{= 96-3}$

$\Rightarrow {{\text{a}}_{\text{24}}}\text{= 93}$

Therefore the required $\text{1}{{\text{7}}^{\text{th}}}$ and ${{18}^{th}}$ terms are $\text{65}$  and $\text{93}$ respectively.

8. Find the ${{\text{7}}^{\text{th}}}$ term in the following sequence whose ${{\text{n}}^{\text{th}}}$ term is ${{\text{a}}_{\text{n}}}\text{=}\frac{{{\text{n}}^{\text{2}}}}{{{\text{2}}^{\text{n}}}}$

Ans:

To find ${{\text{7}}^{\text{th}}}$  term we will substitute $\text{n}$ by $\text{n = 7}$

Here we have 

       ${{\text{a}}_{\text{7}}}\text{=}\frac{{{\text{7}}^{\text{2}}}}{{{\text{2}}^{\text{7}}}}$

      ${{\text{a}}_{\text{7}}}\text{=}\frac{\text{49}}{\text{128}}$

$\Rightarrow {{\text{a}}_{\text{7}}}\text{=}\frac{\text{49}}{\text{128}}$

Therefore the required \[{{\text{7}}^{\text{th}}}\] term ${{\text{a}}_{\text{7}}}\text{=}\frac{\text{49}}{\text{128}}$

9. Find the ${{\text{9}}^{\text{th}}}$ term in the following sequence whose ${{\text{n}}^{\text{th}}}$ term is ${{\text{a}}_{\text{n}}}\text{=(-1}{{\text{)}}^{\text{n-1}}}{{\text{n}}^{\text{3}}}$

Ans:

To find ${{\text{9}}^{\text{th}}}$  term we will substitute $\text{n}$ by $\text{n = 9}$

Here we have 

       \[{{\text{a}}_{\text{9}}}\text{= (-1}{{\text{)}}^{\text{9-1}}}{{\text{(9)}}^{\text{3}}}\]

      ${{\text{a}}_{\text{9}}}\text{=(-1}{{\text{)}}^{\text{8}}}\text{ }\!\!\times\!\!\text{ 729}$

$\Rightarrow {{\text{a}}_{\text{9}}}\text{=729}$

Therefore the required ${{\text{9}}^{\text{th}}}$ term ${{\text{a}}_{\text{9}}}\text{=729}$

10. Find the $\text{2}{{\text{0}}^{\text{th}}}$ term in the following sequence whose ${{\text{n}}^{\text{th}}}$ term is ${{\text{a}}_{\text{n}}}\text{= }\frac{\text{n(n-2)}}{\text{n+3}}$

Ans:

To find $\text{2}{{\text{0}}^{\text{th}}}$  term we will substitute $\text{n}$ by $\text{n = 20}$

Here we have 

       \[{{\text{a}}_{\text{20}}}\text{ = }\frac{\text{20(20-2)}}{\text{20+3}}\]

      ${{\text{a}}_{\text{20}}}\text{ = }\frac{\text{20 }\!\!\times\!\!\text{ 18}}{\text{23}}$

$\Rightarrow {{\text{a}}_{\text{20}}}\text{ = }\frac{\text{360}}{\text{23}}$

Therefore the required $\text{2}{{\text{0}}^{\text{th}}}$ term ${{\text{a}}_{\text{20}}}\text{ = }\frac{\text{360}}{\text{23}}$

11. Find Write the first five terms of the following sequence and obtain the corresponding series: ${{\text{a}}_{\text{1}}}\text{= 3,}$ ${{\text{a}}_{\text{n}}}\text{=3}{{\text{a}}_{\text{n-1}}}\text{+2}$  for all $\text{n  1}$ 

Ans:

To find first five terms of given sequence we will use given values

Here we have 

${{\text{a}}_{\text{1}}}\text{= 3}$ 

Now substituting $\text{n = 2}$ in given sequence we have

     ${{\text{a}}_{2}}\text{=3}{{\text{a}}_{\text{2-1}}}\text{+2}$

$\Rightarrow {{\text{a}}_{\text{2}}}\text{= 3}{{\text{a}}_{\text{1}}}\text{+2}$

$\Rightarrow {{\text{a}}_{\text{2}}}\text{= 3(3)+2}$

$\Rightarrow {{\text{a}}_{\text{2}}}\text{= 11}$

Now substituting $\text{n = 3}$ in given sequence we have

     ${{\text{a}}_{\text{3}}}\text{=3}{{\text{a}}_{\text{3-1}}}\text{+2}$

$\Rightarrow {{\text{a}}_{\text{3}}}\text{= 3}{{\text{a}}_{\text{2}}}\text{+2}$

$\Rightarrow {{\text{a}}_{\text{3}}}\text{= 3(11)+2}$

$\Rightarrow {{\text{a}}_{\text{3}}}\text{= 35}$

Similarly 

     ${{\text{a}}_{4}}\text{= 3}{{\text{a}}_{3}}\text{+2}$

$\Rightarrow {{\text{a}}_{\text{4}}}\text{= 3(35)+2}$

$\Rightarrow {{\text{a}}_{\text{4}}}\text{= 107}$

And 

     ${{\text{a}}_{\text{5}}}\text{= 3}{{\text{a}}_{\text{4}}}\text{+2}$

$\Rightarrow {{\text{a}}_{\text{5}}}\text{= 3(107)+2}$

$\Rightarrow {{\text{a}}_{\text{5}}}\text{= 323}$

Therefore the first five terms of the sequence are $\text{3,11,35,107,323}\text{.}$ 

And the corresponding series is $\text{3+11+35+107+323+}......$

12. Find Write the first five terms of the following sequence and obtain the corresponding series: ${{\text{a}}_{\text{1}}}\text{= -1,}$ ${{\text{a}}_{\text{n}}}\text{=}\frac{{{\text{a}}_{\text{n-1}}}}{\text{n}}$  for all $\text{n}\ge \text{2}$ 

Ans:

To find first five terms of given sequence we will use given values

Here we have 

${{\text{a}}_{\text{1}}}\text{= -1}$ 

Now substituting $\text{n = 2}$ in given sequence we have

     ${{\text{a}}_{\text{2}}}\text{=}\frac{{{\text{a}}_{\text{2-1}}}}{\text{2}}$

$\Rightarrow {{\text{a}}_{\text{2}}}\text{=}\frac{{{\text{a}}_{\text{1}}}}{\text{2}}$

$\Rightarrow {{\text{a}}_{\text{2}}}\text{= -}\frac{\text{1}}{\text{2}}$

Now substituting $\text{n = 3}$ in given sequence we have

     ${{\text{a}}_{\text{3}}}\text{=}\frac{{{\text{a}}_{\text{3-1}}}}{\text{3}}$

\[\Rightarrow {{\text{a}}_{\text{3}}}\text{=}\frac{{{\text{a}}_{\text{2}}}}{\text{3}}\]

$\Rightarrow {{\text{a}}_{\text{3}}}\text{=}\frac{\left( \text{-}\tfrac{\text{1}}{\text{2}} \right)}{\text{3}}$

$\Rightarrow {{\text{a}}_{\text{3}}}\text{= -}\frac{\text{1}}{\text{6}}$

Similarly 

     ${{\text{a}}_{\text{4}}}\text{=}\frac{{{\text{a}}_{\text{4-1}}}}{\text{4}}$

\[\Rightarrow {{\text{a}}_{\text{4}}}\text{=}\frac{{{\text{a}}_{\text{3}}}}{\text{4}}\]

$\Rightarrow {{\text{a}}_{\text{4}}}\text{=}\frac{\left( \text{-}\tfrac{\text{1}}{\text{6}} \right)}{\text{4}}$

$\Rightarrow {{\text{a}}_{\text{4}}}\text{= -}\frac{\text{1}}{\text{24}}$

And 

     ${{\text{a}}_{\text{5}}}\text{=}\frac{{{\text{a}}_{\text{5-1}}}}{\text{5}}$

\[\Rightarrow {{\text{a}}_{\text{5}}}\text{=}\frac{{{\text{a}}_{\text{4}}}}{\text{5}}\]

$\Rightarrow {{\text{a}}_{\text{5}}}\text{=}\frac{\left( \text{-}\tfrac{\text{1}}{\text{24}} \right)}{\text{5}}$

$\Rightarrow {{\text{a}}_{\text{5}}}\text{= -}\frac{\text{1}}{\text{120}}$

Therefore the first five terms of the sequence are $\text{-1,-}\frac{\text{1}}{\text{2}}\text{,-}\frac{\text{1}}{\text{6}}\text{,-}\frac{\text{1}}{\text{24}}\text{,-}\frac{\text{1}}{\text{120}}\text{.}$ 

And the corresponding series is $\left( \text{-1} \right)\text{+}\left( \text{-}\frac{\text{1}}{\text{2}} \right)\text{+}\left( \text{-}\frac{\text{1}}{\text{6}} \right)\text{+}\left( \text{-}\frac{\text{1}}{\text{24}} \right)\text{+}\left( \text{-}\frac{\text{1}}{\text{120}} \right)\text{+}......$

13. Find Write the first five terms of the following sequence and obtain the corresponding series: ${{\text{a}}_{\text{1}}}\text{= }{{\text{a}}_{\text{2}}}\text{= 2,}$ ${{\text{a}}_{\text{n}}}\text{= }{{\text{a}}_{\text{n-1}}}\text{-1}$  for all $\text{n  2}$ 

Ans:

To find first five terms of given sequence we will use given values

Here we have 

${{\text{a}}_{\text{1}}}\text{= 2}$ 

${{\text{a}}_{\text{2}}}\text{= 2}$

Now substituting $\text{n = 3}$ in given sequence we have

     ${{\text{a}}_{\text{3}}}\text{=}{{\text{a}}_{\text{3-1}}}\text{-1}$

$\Rightarrow {{\text{a}}_{\text{3}}}\text{= }{{\text{a}}_{\text{2}}}\text{-1}$

\[\Rightarrow {{\text{a}}_{\text{3}}}\text{= 2-1}\]

$\Rightarrow {{\text{a}}_{\text{3}}}\text{= 1}$

Similarly 

     ${{\text{a}}_{\text{4}}}\text{= }{{\text{a}}_{\text{3}}}\text{-1}$

$\Rightarrow {{\text{a}}_{\text{4}}}\text{= 1-1}$

$\Rightarrow {{\text{a}}_{\text{4}}}\text{= 0}$

And 

     ${{\text{a}}_{\text{5}}}\text{= }{{\text{a}}_{\text{4}}}\text{-1}$

$\Rightarrow {{\text{a}}_{\text{5}}}\text{= 0-1}$

$\Rightarrow {{\text{a}}_{\text{5}}}\text{= -1}$

Therefore the first five terms of the sequence are $\text{2,2,1,0,-1}\text{.}$ 

And the corresponding series is $\text{2 + 2 + 1 + 0 + (-1) + }\text{. }\text{. }\text{.}$

14. The Fibonacci sequence is defined by ${{\text{a}}_{\text{1}}}\text{= }{{\text{a}}_{\text{2}}}\text{= 1}$ and c $\text{n  2}$  

Find $\frac{{{\text{a}}_{\text{n+1}}}}{{{\text{a}}_{\text{n}}}}\text{,}$ for $\text{n=1,2,3,4,5}\text{.}$ 

Ans:

To find the required sequence first we will find the first six terms of given sequence 

Here we have 

${{\text{a}}_{\text{1}}}\text{= 2}$ 

${{\text{a}}_{\text{2}}}\text{= 2}$

Now     ${{\text{a}}_{\text{3}}}\text{=}{{\text{a}}_{\text{2}}}\text{+}{{\text{a}}_{\text{1}}}$ 

$\Rightarrow {{\text{a}}_{\text{3}}}\text{=1+1}$ 

$\Rightarrow {{\text{a}}_{\text{3}}}\text{=2}$

Similarly 

$\Rightarrow {{\text{a}}_{\text{4}}}\text{=2+1}$

$\Rightarrow {{\text{a}}_{\text{4}}}\text{=3}$

And 

$\Rightarrow {{\text{a}}_{\text{5}}}\text{= 3+2}$

$\Rightarrow {{\text{a}}_{\text{5}}}\text{=5}$

And 

$\Rightarrow {{\text{a}}_{\text{6}}}\text{= 5+3}$

\[\Rightarrow {{\text{a}}_{\text{6}}}\text{= 8}\]

Now $\frac{{{\text{a}}_{\text{n+1}}}}{{{\text{a}}_{\text{n}}}}\text{,}$

For $\text{n=1}$ 

$\frac{{{\text{a}}_{\text{2}}}}{{{\text{a}}_{\text{1}}}}\text{=}\frac{\text{1}}{\text{1}}$

$\Rightarrow \frac{{{\text{a}}_{\text{2}}}}{{{\text{a}}_{\text{1}}}}\text{=1}$

For $\text{n=2}$

$\frac{{{\text{a}}_{\text{3}}}}{{{\text{a}}_{\text{2}}}}\text{=}\frac{\text{2}}{\text{1}}$

$\Rightarrow \frac{{{\text{a}}_{\text{3}}}}{{{\text{a}}_{\text{2}}}}\text{=2}$

For $\text{n=3}$

$\frac{{{\text{a}}_{\text{4}}}}{{{\text{a}}_{\text{3}}}}\text{=}\frac{\text{3}}{\text{2}}$

For $\text{n=4}$

$\frac{{{\text{a}}_{\text{5}}}}{{{\text{a}}_{\text{4}}}}\text{=}\frac{\text{5}}{\text{3}}$

For $\text{n=5}$

$\frac{{{\text{a}}_{\text{6}}}}{{{\text{a}}_{\text{5}}}}\text{=}\frac{\text{8}}{\text{5}}$

Hence required $\frac{{{\text{a}}_{\text{n+1}}}}{{{\text{a}}_{\text{n}}}}\text{,}$for \[\text{n=1,2,3,4,5}\text{.}\]are $\text{1,2,}\frac{\text{3}}{\text{2}}\text{,}\frac{\text{5}}{\text{3}}\text{,}\frac{\text{8}}{\text{5}}\text{.}$

NCERT Solutions For Class 11 Maths Chapter 9 Sequences And Series Exercise 9.1

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