NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series - Exercise 9.3

Exercise 9.3

1. Find the $\mathbf{2}{{\mathbf{0}}^{\mathbf{th}}}$ and ${{\mathbf{n}}^{\mathbf{th}}}$ terms of the G.P. $\dfrac{\mathbf{5}}{\mathbf{2}}\mathbf{,}\dfrac{\mathbf{5}}{\mathbf{4}}\mathbf{,}\dfrac{\mathbf{5}}{\mathbf{8}}\mathbf{,}...$.

Ans:

The G.P. is provided as $\dfrac{\text{5}}{\text{2}}\text{,}\dfrac{\text{5}}{\text{4}}\text{,}\dfrac{\text{5}}{\text{8}}\text{,}...$.

Clearly, $a=\dfrac{5}{2}$ is the first term and $r=\dfrac{\dfrac{5}{4}}{\dfrac{5}{2}}=\dfrac{1}{2}$ is the common ratio of the G.P.

Therefore, the ${{20}^{th}}$ term of the G.P. is

$\begin{align}   & {{a}_{20}}=a{{r}^{20-1}} \\  & =\dfrac{5}{2}{{\left( \dfrac{1}{2} \right)}^{19}} \\   & =\dfrac{5}{2\times {{2}^{19}}} \\   & =\dfrac{5}{{{2}^{20}}}. \\ \end{align}$ 

Also, the ${{n}^{th}}$ term of the G.P. is 

$\begin{align}   & {{a}_{n}}=a{{r}^{n-1}} \\  & =\dfrac{5}{2}{{\left( \dfrac{1}{2} \right)}^{n-1}} \\  & =\dfrac{5}{2\cdot {{2}^{n-1}}} \\  & =\dfrac{5}{{{2}^{n}}}. \\ \end{align}$

Hence, the ${{20}^{th}}$ and ${{n}^{th}}$ terms of the G.P. are respectively $\dfrac{5}{{{2}^{20}}}$ and $\dfrac{5}{{{2}^{n}}}$.

2. Find the $\mathbf{1}{{\mathbf{2}}^{\mathbf{th}}}$ term of the G.P. whose ${{\mathbf{8}}^{\mathbf{th}}}$ term is $\mathbf{192}$ and the common ratio is $\mathbf{2}$.

Ans:

The G.P. described in the problem has the common ratio $r=2$.

Suppose that the first term of the G.P. is $a$.

Then, the ${{8}^{th}}$ term of the G.P. is given by

${{a}_{8}}=a{{r}^{8-1}}$$=a{{r}^{7}}$

$\Rightarrow a{{r}^{7}}=192$

$\Rightarrow a{{\left( 2 \right)}^{7}}=192$

$\Rightarrow a{{\left( 2 \right)}^{7}}={{\left( 2 \right)}^{6}}\times 3$

$\Rightarrow a=\dfrac{{{2}^{6}}\times 3}{{{2}^{7}}}=\dfrac{3}{2}$.

Thus, the ${{12}^{th}}$ term,

$\begin{align}   & {{a}_{12}}=a{{r}^{12-1}} \\  & =a{{r}^{11}} \\  & =\dfrac{3}{2}{{\left( 2 \right)}^{11}} \\  & =3{{\left( 2 \right)}^{10}} \\  & =3072. \\ \end{align}$

Hence, the ${{12}^{th}}$ term of the G.P. is $3072$.

3. The ${{\mathbf{5}}^{\mathbf{th}}}\mathbf{,}\,{{\mathbf{8}}^{\mathbf{th}}}\mathbf{,}$ and $\mathbf{1}{{\mathbf{1}}^{\mathbf{th}}}$ terms of a G.P. are $\mathbf{p,}\,\,\mathbf{q,}$ and $\mathbf{s}$ respectively. Show that ${{\mathbf{q}}^{\mathbf{2}}}\mathbf{=ps}$.

Ans:

Suppose that $a$ and $r$ are the first term and the common ratio of the given G.P. Then, by the condition provided to us,

the ${{5}^{th}}$ term of G.P.,

${{a}_{5}}=a{{r}^{5-1}}=a{{r}^{4}}=p$                                  …… (i)

The ${{8}^{th}}$ term of the G.P.,

${{a}_{8}}=a{{r}^{8-1}}=a{{r}^{7}}=q$                                  …… (ii)

Also, the ${{11}^{th}}$  term of the G.P.,

${{a}_{11}}=a{{r}^{11-1}}=a{{r}^{10}}=s$                                …… (iii)

Now, divide the equation (ii) by the equation (i). Then, it gives

$\dfrac{a{{r}^{7}}}{a{{r}^{4}}}=\dfrac{q}{p}$

$\Rightarrow {{r}^{3}}=\dfrac{q}{p}$                                               …… (iv)

Again, divide the equation (iii) by the equation (ii). Then, we have

$\dfrac{a{{r}^{10}}}{a{{r}^{7}}}=\dfrac{s}{q}$

$\Rightarrow {{r}^{3}}=\dfrac{s}{q}$                                           …… (v)

Therefore, from the equation (iv) and (v) gives

$\dfrac{q}{p}=\dfrac{s}{q}$

$\Rightarrow {{q}^{2}}=ps$.

Hence, the required result has been shown.

4. The ${{\mathbf{4}}^{\mathbf{th}}}$ term of a G.P. is a square of its second term, and the first term is $\mathbf{-3}$. Determine its ${{\mathbf{7}}^{\mathbf{th}}}$ term.

Ans:

Suppose that $r$ is the common ratio of the G.P.

It is given that the first term $a=-3$.

Now, by the formula, ${{n}^{th}}$ term of a G.P.,
${{a}_{n}}=a{{r}^{n-1}}$.

Therefore,

${{a}_{4}}=a{{r}^{3}}=\left( -3 \right){{r}^{3}}$ and ${{a}_{2}}=a{{r}^{1}}=\left( -3 \right)r$.

So, by the condition provided to us,

$\left( -3 \right){{r}^{3}}={{\left[ \left( -3 \right)r \right]}^{2}}$

$\Rightarrow -3{{r}^{3}}=9{{r}^{2}}$

$\Rightarrow r=-3$.

Thus, the ${{7}^{th}}$ term of the G.P. is given by

$\begin{align}   & {{a}_{7}}=\left( -3 \right){{\left( -3 \right)}^{7-1}} \\  & ={{\left( -3 \right)}^{7}} \\  & =-2187.  \end{align}$

Hence, the ${{7}^{th}}$ term of the geometric progression is $-2187$.

5. Which term of the following sequences:

(a) $\mathbf{2,2}\sqrt{\mathbf{2}}\mathbf{,4,}...$ is $\mathbf{128}$?

Ans:

The sequence provided to us is in a G.P. since the first term is $a=2$ and the common ratio is $r=\dfrac{2\sqrt{2}}{2}=\sqrt{2}$.

So, suppose that the ${{n}^{th}}$ term of the G.P. is $128$.

Therefore, we have

$2{{\left( \sqrt{2} \right)}^{n-1}}=128$

$\Rightarrow {{2}^{\dfrac{n-1}{2}+1}}={{2}^{7}}$

$\Rightarrow \dfrac{n-1}{2}+1=7$

$\Rightarrow \dfrac{n+1}{2}=6$

$\Rightarrow n+1=12$

$\Rightarrow n=11$.

Hence, $128$ is the ${{11}^{th}}$ term of the G.P. (sequence) $\text{2,2}\sqrt{\text{2}}\text{,4,}...$.

(b) $\sqrt{\mathbf{3}}\mathbf{,3,3}\sqrt{\mathbf{3}}\mathbf{,}...$ is $\mathbf{729}$?

Ans:

The sequence provided to us is in a G.P. since the first term is $a=\sqrt{3}$, and the common ratio is $r=\dfrac{3}{\sqrt{3}}=\sqrt{3}$.

Now, suppose that the ${{n}^{th}}$ term of the G.P. is $729$.

Therefore, we have

$a{{r}^{n-1}}=729$

$\Rightarrow \left( \sqrt{3} \right){{\left( \sqrt{3} \right)}^{\cdot n-1}}=729$

$\Rightarrow {{3}^{\dfrac{1}{2}+\dfrac{n-1}{2}}}={{3}^{6}}$

$\Rightarrow \dfrac{1}{2}+\dfrac{n-1}{2}=6$

$\Rightarrow \dfrac{1+n-1}{2}=6$

$\Rightarrow n=12$.

Hence, $729$ is the ${{n}^{th}}$ term of the geometric progression (sequence) $\sqrt{\text{3}}\text{,3,3}\sqrt{\text{3}}\text{,}...$.

(c) $\dfrac{\mathbf{1}}{\mathbf{3}}\mathbf{,}\dfrac{\mathbf{1}}{\mathbf{9}}\mathbf{,}\dfrac{\mathbf{1}}{\mathbf{27}}\mathbf{,}...$ is $\dfrac{\mathbf{1}}{\mathbf{19683}}$?

Ans:

The sequence given to us is in a geometric progression (G.P.) since the first term of it is $a=\dfrac{1}{3}$ and the common ratio is

$r=\dfrac{\dfrac{1}{9}}{\dfrac{1}{3}}=\dfrac{1}{3}$.

Now, suppose that $\dfrac{1}{19683}$ is the ${{n}^{th}}$ term of the G.P.

Therefore, we have

$a{{r}^{n-1}}=\dfrac{1}{19683}$

$\Rightarrow \dfrac{1}{3}{{\left( \dfrac{1}{3} \right)}^{n-1}}=\dfrac{1}{19683}$

$\Rightarrow {{\left( \dfrac{1}{3} \right)}^{n}}={{\left( \dfrac{1}{3} \right)}^{9}}$

$\Rightarrow n=9$.

Hence, $\dfrac{1}{19683}$ is the ${{9}^{th}}$ term of the G.P. (sequence) $\dfrac{\text{1}}{\text{3}}\text{,}\dfrac{\text{1}}{\text{9}}\text{,}\dfrac{\text{1}}{\text{27}}\text{,}...$.

6. For what values of $\mathbf{x}$, the numbers $\mathbf{-}\dfrac{\mathbf{2}}{\mathbf{7}}\mathbf{,}\,\,\mathbf{x,}\,\,\mathbf{-}\dfrac{\mathbf{7}}{\mathbf{2}}$ are in G.P.?

Ans:

The numbers provided to us are $-\dfrac{2}{7},\,x,\,-\dfrac{7}{2}$.

Now, let ${{a}_{1}}=-\dfrac{2}{7}$, ${{a}_{2}}=x$ and ${{a}_{3}}=-\dfrac{7}{2}$.

Then, $\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{x}{-\dfrac{2}{7}}=\dfrac{-7x}{2}$ and $\dfrac{{{a}_{3}}}{{{a}_{2}}}=\dfrac{-\dfrac{7}{2}}{x}=-\dfrac{7}{2x}$.

Now, if the numbers are in G.P., then

$r=\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}$

$\Rightarrow \dfrac{-7x}{2}=-\dfrac{7}{2x}$

$\Rightarrow {{x}^{2}}=\dfrac{-2\times 7}{-2\times 7}=1$

$\Rightarrow x=\sqrt{1}$

$\Rightarrow x=\pm 1$.

Hence, the numbers $-\dfrac{2}{7},\,x,\,-\dfrac{7}{2}$  are in G.P. if $x=\pm 1$.

7. Find the sum of the numbers $\mathbf{0}\mathbf{.15,}\,\,\mathbf{0}\mathbf{.015,}\,\,\mathbf{0}\mathbf{.0015,}\,...$ in the G.P. up to $\mathbf{20}$ terms.

Ans:

It is provided that the numbers $\text{0}\text{.15,}\,\,\text{0}\text{.015,}\,\,\text{0}\text{.0015,}\,...$are in geometric progression.

Note that, the first term of the sequence is $a=0.15$ and the common ratio is $r=\dfrac{0.015}{0.15}=0.1$

Now, by the formula, sum of $n$ terms in G.P., we have

${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$

Therefore,

${{S}_{20}}=\dfrac{0.15\left[ 1-{{\left( 0.1 \right)}^{20}} \right]}{1-0.1}$

$=\dfrac{0.15}{0.9}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]$

$=\dfrac{15}{90}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]$

$=\dfrac{1}{6}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]$.

8. Find the sum of the numbers $\sqrt{\mathbf{7}}\mathbf{,}\sqrt{\mathbf{21}}\mathbf{,3}\sqrt{\mathbf{7}}\mathbf{,}...$ in the G.P. up to $\mathbf{n}$ terms.

Ans:

It is provided to us that the numbers $\sqrt{7},\sqrt{21},3\sqrt{7},...$ are in geometric progression.

Note that, the first term of the G.P. is $a=\sqrt{7}$ and the common ratio is $r=\dfrac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$.

Now, by the formula, sum of the first $n$ terms in G.P.,

${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$

$\begin{align}   & =\dfrac{\sqrt{7}\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{1-\sqrt{3}} \\  & =\dfrac{\sqrt{7}\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{1-\sqrt{3}}\cdot \dfrac{1+\sqrt{3}}{1+\sqrt{3}} \\  & =\dfrac{\sqrt{7}\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]\left( 1+\sqrt{3} \right)}{1-3} \\  & =\dfrac{-\sqrt{7}\left( \sqrt{3}+1 \right)\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{2} \\ \end{align}$

Hence, the sum of the first $n$ terms is given by

${{S}_{n}}=\dfrac{\sqrt{7}\left( 1+\sqrt{3} \right)}{2}\left[ {{\left( 3 \right)}^{\dfrac{n}{2}}}-1 \right]$.

9. Find the sum of the numbers $\mathbf{1,}\,\,\mathbf{-a,}\,\,{{\mathbf{a}}^{\mathbf{2}}}\mathbf{,}\,\mathbf{-}{{\mathbf{a}}^{\mathbf{3}}}\mathbf{,}...$ (if $\mathbf{a}\ne \mathbf{-1}$) in geometric progression up to $\mathbf{n}$ terms.

Ans:

The numbers $1,-a,{{a}^{2}},-{{a}^{3}},...$ provided to us, is in G.P. with the first term $a=1$ and the common ratio $r=\dfrac{-a}{1}=-a$.

Now, by the formula, sum of the first $n$ terms in G.P.,

${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$

$=\dfrac{1\left[ 1-{{\left( -a \right)}^{n}} \right]}{1-\left( -a \right)}$

Hence, the sum of $n$ terms of the sequence is given by

${{S}_{n}}=\dfrac{\left[ 1-{{\left( -a \right)}^{n}} \right]}{1+a}$.

10. Find the sum of the numbers ${{\mathbf{x}}^{\mathbf{3}}}\mathbf{,}{{\mathbf{x}}^{\mathbf{5}}}\mathbf{,}{{\mathbf{x}}^{\mathbf{7}}}\mathbf{,}...$(if $\mathbf{x}\ne \mathbf{\pm 1}$) in the geometric progression up to $\mathbf{n}$ terms.

Ans:

The numbers ${{x}^{3}},\,{{x}^{5}},\,{{x}^{7}},...$ are in G.P. with its first term $a={{x}^{3}}$ and the common ratio $r=\dfrac{{{x}^{5}}}{{{x}^{3}}}={{x}^{2}}$.

Therefore, by the formula, sum of the first $n$ terms in G.P.,

${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$

$=\dfrac{{{x}^{3}}\left[ 1-{{\left( {{x}^{2}} \right)}^{n}} \right]}{1-{{x}^{2}}}$

$=\dfrac{{{x}^{3}}\left( 1-{{x}^{2n}} \right)}{1-{{x}^{2}}}$.

Hence, the sum of the first $n$ terms of the sequence ${{\text{x}}^{\text{3}}}\text{,}{{\text{x}}^{\text{5}}}\text{,}{{\text{x}}^{\text{7}}}\text{,}...$ is given by  ${{S}_{n}}=\dfrac{{{x}^{3}}\left( 1-{{x}^{2n}} \right)}{1-{{x}^{2}}}$.

11. Evaluate $\sum\limits_{\mathbf{k=1}}^{\mathbf{11}}{\left( \mathbf{2+}{{\mathbf{3}}^{\mathbf{k}}} \right)}$.

Ans:

The given sum can be written as

$\sum\limits_{\text{k=1}}^{\text{11}}{\left( \text{2+}{{\text{3}}^{\text{k}}} \right)}=\sum\limits_{k=1}^{11}{2}+\sum\limits_{\text{k=1}}^{\text{11}}{{{\text{3}}^{\text{k}}}}=22+\sum\limits_{\text{k=1}}^{\text{11}}{{{\text{3}}^{\text{k}}}}$                    …… (i)

Again, $\sum\limits_{\mathbf{k=1}}^{\mathbf{11}}{{{3}^{k}}}={{3}^{1}}+{{3}^{2}}+{{3}^{3}}+\cdot \cdot \cdot +{{3}^{11}}$, which is in a geometric progression with the first term $a=3$ and the common ratio $r=3$.

Therefore, we have

${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$

$\Rightarrow {{S}_{n}}=\dfrac{3\left[ {{3}^{11}}-1 \right]}{3-1}$

$\Rightarrow {{S}_{n}}=\dfrac{3}{2}\left( {{3}^{11}}-1 \right)$

Thus,

$\sum\limits_{k=1}^{11}{{{3}^{k}}}=\dfrac{3}{2}\left( {{3}^{11}}-1 \right)$.                                                 …… (ii)

Hence, from the equation (i) and (ii), it gives

$\sum\limits_{\text{k=1}}^{\text{11}}{\left( \text{2+}{{\text{3}}^{\text{k}}} \right)}=22+\dfrac{3}{2}\left( {{3}^{11}}-1 \right)$.

12. The sum of the first three terms of a G.P. is $\dfrac{\mathbf{39}}{\mathbf{10}}$ and their product is $\mathbf{1}$. Find the common ratio and the terms.

Ans:

Suppose that the first three terms of the geometric progression are $\dfrac{a}{r},\,\,a,\,\,ar$.

Now, by the given conditions,

$\dfrac{a}{r}+a+ar=\dfrac{39}{10}$                                        …… (i)

$\left( \dfrac{a}{r} \right)\cdot \left( a \right)\cdot \left( ar \right)=1$                                    …… (ii)

Simplifying the equation (ii) gives

${{a}^{3}}=1$

$\Rightarrow a=1$, neglecting the imaginary roots.

Thus, replacing $a$ by $1$ in the equation (i) gives

$\dfrac{1}{r}+1+r=\dfrac{39}{10}$

$\Rightarrow 1+r+{{r}^{2}}=\dfrac{39}{10}r$

$\Rightarrow 10+10r+10{{r}^{2}}-39r=0$

$\Rightarrow 10{{r}^{2}}-29r+10=0$

$\Rightarrow 10{{r}^{2}}-25r-4r+10=0$

$\Rightarrow \left( 5r-2 \right)\left( 2r-5 \right)=0$

$\Rightarrow r=\dfrac{2}{5}$ or $\dfrac{5}{2}$.

Hence, $\dfrac{5}{2},$ $1$, and $\dfrac{2}{5}$ are the three terms of the geometric progression.

13. How many terms of G.P. $\mathbf{3},\,{{\mathbf{3}}^{\mathbf{2}}},\,{{\mathbf{3}}^{\mathbf{3}}},...$ are needed to give the sum $\mathbf{120}$.

Ans:

It is provided that the terms $\text{3,}\,{{\text{3}}^{\text{2}}}\text{,}\,{{\text{3}}^{\text{3}}}\text{,}...$ are geometric progression with the first term $a=3$ and the common ratio $r=3$.

Suppose that to get the sum of the G.P. as $120$, $n$ terms are needed to be considered.

So, by the formula, sum of $n$ terms in G.P.,

${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$.

Therefore,

$\begin{align}  & \dfrac{3\left( {{3}^{n}}-1 \right)}{3-1}=120 \\  & \Rightarrow {{3}^{n}}-1=120\times \dfrac{2}{3} \\  & \Rightarrow {{3}^{n}}-1=80 \\  & \Rightarrow {{3}^{n}}=81 \\  & \Rightarrow {{3}^{n}}={{3}^{4}} \\ \end{align}$

$\Rightarrow n=4$.

Hence, to get the sum of $\text{3,}\,{{\text{3}}^{\text{2}}}\text{,}\,{{\text{3}}^{\text{3}}}\text{,}...$ as $120$, $4$ terms are needed to be considered.

14. The sum of the first three terms of a G.P. is $\mathbf{16}$ and the sum of the next three terms is $\mathbf{128}$. Determine the first term, the common ratio and the sum of $\mathbf{n}$ terms of the G.P.

Ans:

Suppose that the first six terms of the given geometric progression are

$a,\,ar,\,a{{r}^{2}},\,a{{r}^{3}},\,a{{r}^{4}},\,a{{r}^{5}}$.

Then, by the condition provided to us,

$a+ar+a{{r}^{2}}=16$              

 $\Rightarrow a\left( 1+r+{{r}^{3}} \right)=16$                                           …… (i)

and $a{{r}^{3}}+a{{r}^{4}}+a{{r}^{5}}=128$

$\Rightarrow a{{r}^{3}}\left( 1+r+{{r}^{2}} \right)=128$                                       …… (ii)

Now, divide the equation (ii) by the equation (i). Then, it gives

$\dfrac{a{{r}^{3}}\left( 1+r+{{r}^{2}} \right)}{a\left( 1+r+{{r}^{2}} \right)}=\dfrac{128}{16}$

$\begin{align}   & \Rightarrow {{r}^{3}}=8 \\  & \Rightarrow {{r}^{3}}={{2}^{3}} \\ \end{align}$

$\Rightarrow r=2$                                                            …… (iii)

Therefore, the equation (i) and (iii) together implies

$a\left( 1+2+{{2}^{2}} \right)=16$

$\begin{align}   & \Rightarrow a\times 7=16 \\  & \Rightarrow a=\dfrac{16}{7} \\ \end{align}$

Hence, the sum of $n$ terms of the G.P. is given by

${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$

$=\dfrac{16}{7}\dfrac{\left( {{2}^{n}}-1 \right)}{2-1}$

$=\dfrac{16}{7}\left( {{2}^{n}}-1 \right)$.

15. Given a G.P. with $\mathbf{a}=\mathbf{729}$ and the ${{\mathbf{7}}^{\mathbf{th}}}$ term $\mathbf{64}$, determine ${{\mathbf{S}}_{\mathbf{7}}}$.

Ans:

The first term of the G.P. is $a=729$.

Suppose that $r$ is the common ratio of the geometric progression.

Therefore, by the formula, ${{n}^{th}}$ term of the sequence in G.P.,

${{a}_{n}}=a{{r}^{n-1}}$

So, ${{a}_{7}}=\left( 729 \right){{r}^{7-1}}=729\cdot {{r}^{6}}$

$\begin{align}   & \Rightarrow 64=729{{r}^{6}} \\  & \Rightarrow {{r}^{6}}={{\left( \dfrac{2}{3} \right)}^{6}} \\  & \Rightarrow r=\dfrac{2}{3}. \\ \end{align}$

Thus, by the formula, sum of $n$ terms in G.P.,

${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$

Therefore, for $n=7$, we have

${{S}_{7}}=\dfrac{729\left[ 1-{{\left( \dfrac{2}{3} \right)}^{7}} \right]}{1-\dfrac{2}{3}}$

$={{3}^{7}}\times \dfrac{{{3}^{7}}-{{2}^{7}}}{{{3}^{7}}}$

$\begin{align}   & ={{3}^{7}}-{{2}^{7}} \\  & =2187-128 \\ \end{align}$

$=2059$.

Hence, ${{S}_{7}}=2059$.

16. Find a G.P. for which the sum of the first two terms is $-\mathbf{4}$ and the fifth term is $\mathbf{4}$ times the third term.

Ans:

Suppose that $a$ and $r$ respectively are the first term and the common ratio of the geometric progression.

So, by the condition provided to us,

$-4=\dfrac{a\left( 1-{{r}^{2}} \right)}{1-r}$                                                    …… (i)

${{a}_{5}}=4\times {{a}_{3}}$                                                          …… (ii)

Now, we know that, the ${{n}^{th}}$ term of a G.P.,

${{a}_{n}}=a{{r}^{n-1}}$.
Therefore, 

${{a}_{3}}=a{{r}^{3-1}}=a{{r}^{2}}$                                                   …… (iii)

and ${{a}_{5}}=a{{r}^{5-1}}=a{{r}^{4}}$.                                           …… (iv)

Thus, from the equation (ii), (iii), and (iv) we have

$a{{r}^{4}}=4\times a{{r}^{2}}$

$\Rightarrow {{r}^{2}}=4$

$\Rightarrow r=\pm 2$.

Substituting $r=2$ into the equation (i), it gives

$-4=\dfrac{a\left[ 1-{{\left( 2 \right)}^{2}} \right]}{1-2}$

$\Rightarrow -4=\dfrac{a\left( 1-4 \right)}{-1}$

$\Rightarrow -4=3a$

$\Rightarrow a=-\dfrac{4}{3}$.

Also, substituting $r=-2$ into the equation (i), gives

$-4=\dfrac{a\left[ 1-{{\left( -2 \right)}^{2}} \right]}{1-\left( -2 \right)}$

$\Rightarrow -4=\dfrac{a\left( 1-4 \right)}{1+2}$

$\Rightarrow -4=\dfrac{a\left( -3 \right)}{3}$

$\Rightarrow a=4$.

Hence, the G.P. can be of the form $-\dfrac{4}{3},-\dfrac{8}{3},-\dfrac{16}{3},...$ or $4,-8,-16,-32,...$

17. If the ${{\mathbf{4}}^{\mathbf{th}}}$ , $\mathbf{1}{{\mathbf{0}}^{\mathbf{th}}}$, and $\mathbf{1}{{\mathbf{6}}^{\mathbf{th}}}$ terms of a G.P. are $\mathbf{x,y}$ and $\mathbf{z}$, respectively. Prove that $\mathbf{x},\,\mathbf{y},$$\mathbf{z}$ are in G.P.

Ans:

Suppose that $a$ and $r$ respectively are the first term and common ratio of the geometric progression.

Then, by the condition provided to us,

${{a}_{4}}=a{{r}^{3}}=x$                                             …… (i)

${{a}_{10}}=a{{r}^{9}}=y$                                            …… (ii)

${{a}_{16}}=a{{r}^{15}}=z$                                           …… (iii)

Now, divide the equation (ii) by the equation (i). Then, it gives

$\dfrac{y}{x}=\dfrac{a{{r}^{9}}}{a{{r}^{3}}}$

$\Rightarrow \dfrac{y}{x}={{r}^{6}}$                                                  ……(iv)

Again, divide the equation (iii) by the equation (ii). Then, it gives

$\dfrac{z}{y}=\dfrac{a{{r}^{15}}}{a{{r}^{9}}}$

$\Rightarrow \dfrac{z}{y}={{r}^{6}}$                                                  ..,…. (v)

Equating the equations (iv) and (v), we have

$\dfrac{y}{x}=\dfrac{z}{y}$.

Hence, it has been proved that $x,y,z$ are in G.P.

18. Find the sum of $\mathbf{n}$ terms of the sequence $\mathbf{8},\,\mathbf{88},\,\mathbf{888},\,\mathbf{8888},...$.

Ans:

The provided sequence $\text{8,}\,\text{88,}\,\text{888,}\,\text{8888,}...$ is not in a geometric progression (G.P.), but by rearranging and rewriting the terms, it can be converted into a G.P.

The sum of the $n$ terms of the given sequence is given by

${{S}_{n}}=8+88+888+8888+\cdot \cdot \cdot \,\text{to n terms}$

$=\dfrac{8}{9}\left[ 9+99+999+9999+\cdot \cdot \cdot \,\text{to n terms} \right]$

$=\dfrac{8}{9}\left[ \left( 10-1 \right)+\left( {{10}^{2}}-1 \right)+\left( {{10}^{3}}-1 \right)+\left( {{10}^{4}}-1 \right)+\cdot \cdot \cdot \,\,\text{to n terms} \right]$

$=\dfrac{8}{9}\left[ \left( 10+{{10}^{2}}+\cdot \cdot \cdot +\,\,\text{n terms} \right)-\left( 1+1+1+\cdot \cdot \cdot \,\text{n terms} \right) \right]$

$\begin{align}   & =\dfrac{8}{9}\left[ \dfrac{10\left( {{10}^{n}}-1 \right)}{10-1}-n \right] \\  & =\dfrac{80}{81}\left( {{10}^{n}}-1 \right)-\dfrac{8}{9}n. \\  \end{align}$

Hence, the sum of $n$ terms of the sequence $\text{8,}\,\text{88,}\,\text{888,}\,\text{8888,}...$ is $\dfrac{80}{81}\left( {{10}^{n}}-1 \right)-\dfrac{8}{9}n$.

19. Find the sum of the products of the corresponding terms of the sequences $\mathbf{2},\,\mathbf{4},\,\mathbf{8},\,\mathbf{16},\,\mathbf{32}$ and $\mathbf{128},\,\mathbf{32},\,\mathbf{8},\,\mathbf{2},\,\dfrac{\mathbf{1}}{\mathbf{2}}$.

Ans:

The sum of the product of the corresponding terms $\text{2,}\,\text{4,}\,\text{8,}\,\text{16,}\,\text{32}$ and $\text{128,}\,\text{32,}\,\text{8,}\,\text{2,}\,\dfrac{\text{1}}{\text{2}}$, is given by

$=2\times 128+4\times 32+8\times 8+16\times 2+32\times \dfrac{1}{2}$

$=64\left[ 4+2+1+\dfrac{1}{2}+\dfrac{1}{{{2}^{2}}} \right]$.

Now, note that, the terms $4,\,2,\,1,\,\dfrac{1}{2},\,\dfrac{1}{{{2}^{2}}}$ are in geometric progression, with the first term $a=4$ and the common ratio $r=\dfrac{2}{4}=\dfrac{1}{2}$.

So, by the formula, sum of $n$ terms in G.P.,

${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$, $r<1$.

Therefore,

$\begin{align}  & {{S}_{5}}=\dfrac{4\left[ 1-{{\left( \dfrac{1}{2} \right)}^{5}} \right]}{1-\dfrac{1}{2}} \\  & =\dfrac{4\left[ 1-\dfrac{1}{32} \right]}{\dfrac{1}{2}} \\  & =8\left( \dfrac{32-1}{32} \right) \\  & =\dfrac{31}{4}. \\ \end{align}$

Hence, the sum of the products of the corresponding terms of the given sequences is $\dfrac{31}{4}$.

20. Show that the products of the corresponding terms of the sequences $\mathbf{a,}\,\mathbf{ar,}\,\mathbf{a}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{,}...\mathbf{,}\,\mathbf{a}{{\mathbf{r}}^{\mathbf{n-1}}}$ and $\mathbf{A},\,\mathbf{AR},\,\mathbf{A}{{\mathbf{R}}^{\mathbf{2}}},...,\,\mathbf{A}{{\mathbf{R}}^{\mathbf{n}-\mathbf{1}}}$ form a G.P. and find the common ratio.

Ans:

The product of the corresponding terms of the sequences $\text{a,}\,\text{ar,}\,\text{a}{{\text{r}}^{\text{2}}}\text{,}...\text{,}\,\text{a}{{\text{r}}^{\text{n-1}}}$ and $\text{A,}\,\text{AR,}\,\text{A}{{\text{R}}^{\text{2}}}\text{,}...\text{,}\,\text{A}{{\text{R}}^{\text{n-1}}}$ is $aA,\,arAR,\,a{{r}^{2}}A{{R}^{2}},...,a{{r}^{n-1}}A{{R}^{n-1}}$.

Now, we are to be proved that the product is in G.P.

So. $\dfrac{\text{Second term }}{\text{First term}}=\dfrac{arAR}{aA}=rR$                                  …… (i)

Also,

$\dfrac{\text{Third term}}{\text{Second term}}=\dfrac{a{{r}^{2}}A{{R}^{2}}}{arAR}=rR$                                      …… (ii)

Thus, by equating the equations (i) and (ii), we have

$\dfrac{\text{Second term }}{\text{First term}}=\dfrac{\text{Third term}}{\text{Second term}}$$=rR$, the common ratio.

Hence, it has been proved that the product of the corresponding terms of the sequences $\text{a,}\,\text{ar,}\,\text{a}{{\text{r}}^{\text{2}}}\text{,}...\text{,}\,\text{a}{{\text{r}}^{\text{n-1}}}$ and $\text{A,}\,\text{AR,}\,\text{A}{{\text{R}}^{\text{2}}}\text{,}...\text{,}\,\text{A}{{\text{R}}^{\text{n-1}}}$ is in G.P. with the common ratio $rR$.

21. Find four numbers forming a geometric progression in which the third term is greater than the first term by $\mathbf{9}$, and the second term is greater than the ${{\mathbf{4}}^{\mathbf{th}}}$ by $\mathbf{18}$.

Ans:

Suppose that the first term and the common ratio of the geometric progression respectively are $a$ and $r$.

Then the first four terms of the sequence are $a,\,ar,\,a{{r}^{2}},\,a{{r}^{3}}$.

Now, according to the condition provided,

$a{{r}^{2}}=a+9$                                                    …… (i)

$ar=a{{r}^{3}}+18$                                                 …… (ii)

Therefore, rewriting the equations, give

$a\left( {{r}^{2}}-1 \right)=9$                                                …… (iii)

$ar\left( 1-{{r}^{2}} \right)=18$                                              …… (iv)

Now, divide the equation (iv) by the equation (iii). Then, it gives

$\dfrac{ar\left( 1-{{r}^{2}} \right)}{-a\left( 1-{{r}^{2}} \right)}=\dfrac{18}{9}$

$\Rightarrow r=-2$                                                   …… (v)

Now, using the equation (v) into the equation (i) gives

$\begin{align}   & 4a=a+9 \\  & \Rightarrow 3a=9 \\  & \Rightarrow a=3 \\ \end{align}$

Hence, the first four numbers that forms a geometric progression are $3,\,3\left( -2 \right),\,3{{\left( -2 \right)}^{2}},$ and $3{{\left( -2 \right)}^{3}}$, that is $3,\,-6,\,\,12,\,$ and $-24$.

22. If the ${{\mathbf{p}}^{\mathbf{th}}}$ , ${{\mathbf{q}}^{\mathbf{th}}}$ and ${{\mathbf{r}}^{\mathbf{th}}}$ terms of a G.P. are $\mathbf{a},\,\mathbf{b},\,$ and $\mathbf{c}$ respectively. Prove that ${{\mathbf{a}}^{\mathbf{q-r}}}\,{{\mathbf{b}}^{\mathbf{r-p}}}\,{{\mathbf{c}}^{\mathbf{p-q}}}\mathbf{=1}$.

Ans:

First suppose that the first term and the common ratio of the geometric progression are respectively $A$ and $R$.

Then, by the condition provided to us,

$A{{R}^{p-1}}=a$,                                       …… (i)

$A{{R}^{q-1}}=b$,                                       …… (ii)

and $A{{R}^{r-1}}=c$.                                 …… (iii)

Therefore, we have

${{a}^{q-r}}{{b}^{r-p}}{{c}^{p-q}}$

$={{A}^{q-r}}\times {{R}^{\left( p-1 \right)\left( q-r \right)}}\times {{A}^{r-p}}\times {{R}^{\left( q-1 \right)\left( r-p \right)}}\times {{A}^{p-q}}\times {{R}^{\left( r-1 \right)\left( p-q \right)}}$, using the equations (i), (ii) and (iii).

$\begin{align}   & ={{A}^{q-r+r-p+p-q}}\times {{R}^{\left( pr-pr-q+r \right)\left( rq-r+p-pq \right)+\left( pr-p-qr+q \right)}} \\  & ={{A}^{0}}\times {{R}^{0}} \\  & =1. \\ \end{align}$

Hence, it has been proved that, ${{a}^{q-r}}{{b}^{r-p}}{{c}^{p-q}}=1$.

23. If the first and the ${{\mathbf{n}}^{\mathbf{th}}}$ term of a G.P. are $\mathbf{a}$ and $\mathbf{b}$, respectively, and if $\mathbf{P}$ is the product of $\mathbf{n}$ terms, prove that ${{\mathbf{P}}^{\mathbf{2}}}\mathbf{=}{{\left( \mathbf{ab} \right)}^{\mathbf{n}}}$.

Ans:

First suppose that, $r$ is the common ratio of the geometric progression.

Since, $a$ is the first of the G.P., so it takes the form $a,\,ar,\,a{{r}^{2}},a{{r}^{3}},...,a{{r}^{n-1}}$.

Therefore, by the given condition,

$b=a{{r}^{n-1}}$                                              …… (i)

Also, 

$P=$ Product of $n$ terms

$=\left( a \right)\left( ar \right)\left( a{{r}^{2}} \right)\cdot \cdot \cdot \left( a{{r}^{n-1}} \right)$

$=\left( a\times a\times a\times \cdot \cdot \cdot \times a \right)\left( r\times {{r}^{2}}\times \cdot \cdot \cdot \times {{r}^{n-1}} \right)$

$={{a}^{n}}{{r}^{1+2+\cdot \cdot \cdot +\left( n-1 \right)}}$                                      …… (ii)

Now, notice that the numbers $1,2,...,\left( n-1 \right)$ on the power of $r$ are in A.P.

So, the sum of the $n-1$ terms, 

$1+2+\cdot \cdot \cdot +\left( n-1 \right)=\dfrac{n-1}{2}\left[ 2+\left( n-1-1 \right)\times 1 \right]$

$\begin{align}   & =\dfrac{n-1}{2}\left[ 2+n-2 \right] \\  & =\dfrac{n\left( n-1 \right)}{2}. \\ \end{align}$

Thus, from the equation (ii) we have

$P={{a}^{n}}{{r}^{\dfrac{n\left( n-1 \right)}{2}}}$

$\begin{align}   & \Rightarrow {{P}^{2}}={{a}^{2n}}{{r}^{n\left( n-1 \right)}} \\  & ={{\left[ {{a}^{2}}{{r}^{\left( n-1 \right)}} \right]}^{n}} \\  & ={{\left( a\times a{{r}^{n-1}} \right)}^{n}} \\ \end{align}$

$={{\left( ab \right)}^{n}}$, applying the result of equation (i).

Hence, it has been proved that $P={{\left( ab \right)}^{n}}$.

24. Show that the ratio of the sum of first $\mathbf{n}$ terms of a G.P. to the sum of terms from ${{\left( \mathbf{n+1} \right)}^{\mathbf{th}}}$ to ${{\left( \mathbf{2n} \right)}^{\mathbf{th}}}$ term is $\dfrac{\mathbf{1}}{{{\mathbf{r}}^{\mathbf{n}}}}$.

Ans:

Suppose that $a$ and $r$ respectively are the first term and the common ratio of the geometric progression.

Therefore, by the formula, sum of the first $n$ terms of a G.P.,

${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$.

Now, note that from ${{\left( \text{n+1} \right)}^{\text{th}}}$ to ${{\left( \text{2n} \right)}^{\text{th}}}$, there are $n$ terms.

So, the sum of the terms between these two terms,

${{{S}'}_{n}}=\dfrac{{{a}_{n+1}}\left( 1-{{r}^{n}} \right)}{1-r}$.

Thus, the ${{\left( n+1 \right)}^{th}}$ term, ${{a}_{n+1}}=a{{r}^{n+1-1}}=a{{r}^{n}}$.

So, the ratio of the sums,

$\begin{align}   & \dfrac{{{S}_{n}}}{{{{{S}'}}_{n}}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\times \dfrac{\left( 1-r \right)}{a{{r}^{n}}\left( 1-{{r}^{n}} \right)} \\  & =\dfrac{1}{{{r}^{n}}}. \\ \end{align}$

Hence, it has been shown that the ratio of the sum of the sum of first $\text{n}$ terms of a G.P. to the sum of terms from ${{\left( \text{n+1} \right)}^{\text{th}}}$ to ${{\left( \text{2n} \right)}^{\text{th}}}$ term is $\dfrac{\text{1}}{{{\text{r}}^{\text{n}}}}$. 

25. If $\mathbf{a,}\,\mathbf{b,c}$ and $\mathbf{d}$ are in G.P., then show that

$\left( {{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{b}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{c}}^{\mathbf{2}}} \right)\left( {{\mathbf{b}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{d}}^{\mathbf{2}}} \right)\mathbf{=}{{\left( \mathbf{ab+bc+cd} \right)}^{\mathbf{2}}}$.

Ans:

It is given that the numbers $a,b,c$ and $d$ are in geometric progression (G.P.). So, we have

$\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}$.

That is,

$bc=ad$                                    …… (i)

${{b}^{2}}=ac$                                     …… (ii) 

${{c}^{2}}=bd$                                     …… (iii)

Now, Right-Hand-Side of the statement that needs to be proved,

$={{\left( ab+bc+cd \right)}^{2}}$

$={{\left( ab+ad+cd \right)}^{2}}$, applying the result of the equation (i)

 $={{\left[ ab+d\left( a+c \right) \right]}^{2}}$

$={{a}^{2}}{{b}^{2}}+2abd\left( a+c \right)+{{d}^{2}}{{\left( a+c \right)}^{2}}$

$={{a}^{2}}{{b}^{2}}+2{{a}^{2}}bd+2acbd+{{d}^{2}}\left( {{a}^{2}}+2ac+{{c}^{2}} \right)$

\[={{a}^{2}}{{b}^{2}}+2{{a}^{2}}{{c}^{2}}+2{{b}^{2}}{{c}^{2}}+{{d}^{2}}{{a}^{2}}+2{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{c}^{2}}\], applying the results of equations (i) and (ii).

$={{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}+{{d}^{2}}{{a}^{2}}+{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{c}^{2}}$

$={{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}\cdot {{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}+{{c}^{2}}{{b}^{2}}+{{c}^{2}}\times {{c}^{2}}+{{c}^{2}}{{d}^{2}}$, applying the results of equations (ii) and (iii) and rewriting the terms.

$={{a}^{2}}\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)+{{b}^{2}}\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)+{{c}^{2}}\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)$

$=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)$

$=$ Left-Hand-Side of the proof statement.

Hence, it has been proved that

$\left( {{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}} \right)\left( {{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}{{\text{d}}^{\text{2}}} \right)\text{=}{{\left( \text{ab+bc+cd} \right)}^{\text{2}}}$.

26. Insert two numbers between $\mathbf{3}$ and $\mathbf{81}$ so that the resulting sequence is G.P.

Ans:

Suppose the two numbers that needed to be inserted between $3$ and $31$ are ${{G}_{1}}$ and ${{G}_{2}}$, so that it forms a G.P. sequence $3,\,{{G}_{1}},\,{{G}_{2}},\,81$.

Also, assume that the first term and the common ratio respectively are $a$ and $r$.

Then, according to the given condition we have

$3{{\left( r \right)}^{4-1}}=81$

$\Rightarrow {{r}^{3}}=27$

$\Rightarrow r=3$, neglecting the imaginary roots.

Therefore, the terms 

${{G}_{1}}=ar=3\times 3=9$ and 

${{G}_{2}}=a{{r}^{2}}=3{{\left( 3 \right)}^{2}}=27$.

Hence, the numbers that need to be inserted between $3$ and $81$ are $9$ and $27$.

27. Find the value of $\mathbf{n}$ so that $\dfrac{{{\mathbf{a}}^{\mathbf{n+1}}}\mathbf{+}{{\mathbf{b}}^{\mathbf{n+1}}}}{{{\mathbf{a}}^{\mathbf{n}}}\mathbf{+}{{\mathbf{b}}^{\mathbf{n}}}}$ may be the geometric mean between $\mathbf{a}$ and $\mathbf{b}$.

Ans:

It is known that the geometric mean between $a$ and $b$ is $\sqrt{ab}$.

Therefore, according to the condition provided,

$\dfrac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}=\sqrt{ab}$

$\Rightarrow \dfrac{{{\left( {{a}^{n+1}}+{{b}^{n+1}} \right)}^{2}}}{{{\left( {{a}^{n}}+{{b}^{n}} \right)}^{2}}}=ab$, squaring both sides of the equation.

$\Rightarrow {{a}^{2n+2}}+2{{a}^{n+1}}{{b}^{n+1}}+{{b}^{2n+2}}=\left( ab \right)\left( {{a}^{2n}}+2{{a}^{n}}{{b}^{n}}+{{b}^{2n}} \right)$

$\Rightarrow {{a}^{2n+2}}+2{{a}^{n+1}}{{b}^{n+1}}+{{b}^{2n+2}}={{a}^{2n+1}}b+2{{a}^{n+1}}{{b}^{n+1}}+a{{b}^{2n+1}}$

$\Rightarrow {{a}^{2n+2}}+{{b}^{2n+2}}={{a}^{2n+1}}b+a{{b}^{2n+1}}$

$\Rightarrow {{a}^{2n+2}}-{{a}^{2n+1}}b=a{{b}^{2n+1}}-{{b}^{2n+2}}$.

$\Rightarrow {{a}^{2n+1}}\left( a-b \right)={{b}^{2n+1}}\left( a-b \right)$

$\Rightarrow {{\left( \dfrac{a}{b} \right)}^{2n+1}}=1={{\left( \dfrac{a}{b} \right)}^{0}}$

$\Rightarrow 2n+1=0$

$\Rightarrow n=-\dfrac{1}{2}$.

Hence, for $n=-\dfrac{1}{2}$, $\dfrac{{{\text{a}}^{\text{n-1}}}\text{+}{{\text{b}}^{\text{n-1}}}}{{{\text{a}}^{\text{n}}}\text{+}{{\text{b}}^{\text{n}}}}$ may be the geometric mean between $\text{a}$ and $b$.

28. The sum of two numbers is $\mathbf{6}$ times their geometric mean, show that numbers are in the ratio $\left( \mathbf{3+2}\sqrt{\mathbf{2}} \right)\mathbf{:}\left( \mathbf{3-2}\sqrt{\mathbf{2}} \right)$.

Ans:

Suppose that $a$ and $b$ are the two numbers.

Then, their geometric mean $=\sqrt{ab}$.

So, by the condition provided to us,

$a+b=6\sqrt{ab}$                                                  …… (i)

Squaring both sides of the equation (i), we have

${{\left( a+b \right)}^{2}}=36ab$                                             …… (ii)

Now,

${{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab$

$=36ab-4ab$, using the equation (ii).

$=32ab$.

$\Rightarrow a-b=\sqrt{32}\sqrt{ab}=4\sqrt{2}\sqrt{ab}$                      …… (iii)

Therefore, adding the equation (i) and (iii), gives

$2a=\left( 6+4\sqrt{2} \right)\sqrt{ab}$

$\Rightarrow a=\left( 3+2\sqrt{2} \right)\sqrt{ab}$                                        …… (iv)

Now, using the equation (iv) into the equation (i), gives

$b=6\sqrt{ab}-\left( 3+2\sqrt{2} \right)\sqrt{ab}$

$\Rightarrow b=\left( 3-2\sqrt{2} \right)\sqrt{ab}$                                         ….. (v)

Thus, dividing the equation (iv) by (v) we have,

$\dfrac{a}{b}=\dfrac{\left( 3+2\sqrt{2} \right)\sqrt{ab}}{\left( 3-2\sqrt{2} \right)\sqrt{ab}}=\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}$

Hence, the ratio of the numbers is $\left( 3+2\sqrt{2} \right):\left( 3-2\sqrt{2} \right)$.

29. If $\mathbf{A}$ and $\mathbf{G}$ be A.M. and G.M., respectively between two positive numbers, prove that the numbers are $\mathbf{A\pm }\sqrt{\left( \mathbf{A+G} \right)\left( \mathbf{A-G} \right)}$.

Ans:

Suppose that $a$ and $b$ are the two positive numbers.

Then, according to the definition, A.M and G.M of the numbers are

$A=\dfrac{a+b}{2}$                                 …… (i)

$G=\sqrt{ab}$                                  …… (ii)

Rewriting the equations (i) and (ii), gives

$a+b=2A$                               …… (iii)

$ab={{G}^{2}}$                                   …… (iv)

Now, it is known that,

${{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab$

$={{\left( 2A \right)}^{2}}-4{{G}^{2}}$, using the equations (iii) and (iv).

$\begin{align}   & =4{{A}^{2}}-4{{G}^{2}} \\  & =4\left( {{A}^{2}}-{{G}^{2}} \right) \\ \end{align}$

Therefore,

${{\left( a-b \right)}^{2}}=4\left( A+G \right)\left( A-G \right)$

$\Rightarrow a-b=2\sqrt{\left( A+G \right)\left( A-G \right)}$       …… (v)

Adding the equations (iii) and (v) we have

$2a=2A+2\sqrt{\left( A+G \right)\left( A-G \right)}$

$\Rightarrow a=A+\sqrt{\left( A+G \right)\left( A-G \right)}$        …… (vi)

Thus, the equations (iii) and (vi) gives

$\begin{align}   & b=2A-A-\sqrt{\left( A+G \right)\left( A-G \right)} \\  & =A-\sqrt{\left( A+G \right)\left( A-G \right)}. \\ \end{align}$

Hence, the two positive numbers are $A\pm \sqrt{\left( A+G \right)\left( A-G \right)}$.

30. The number of bacteria in a certain culture doubles every hour. If there were $\mathbf{30}$ bacteria present in the culture originally, how many bacteria will be present at the end of ${{\mathbf{2}}^{\mathbf{nd}}}$ hour, ${{\mathbf{4}}^{\mathbf{th}}}$ hour, and ${{\mathbf{n}}^{\mathbf{th}}}$ hour?

Ans:
Since, the number of bacteria is increasing by two times in each hour, so the problem makes a geometric progression.

It is given that, originally there were $30$ bacteria.

So, the first term of the G.P. is $a=30$ and the common ratio is $r=2$.

Thus, the number of bacteria will present at the end of ${{2}^{nd}}$ hour is given by  ${{a}_{3}}=a{{r}^{3-1}}=\left( 30 \right){{\left( 2 \right)}^{2}}=120$.

The number of bacteria will present at the end of ${{4}^{th}}$ hour is given by

${{a}_{5}}=a{{r}^{5-1}}=\left( 30 \right){{\left( 2 \right)}^{4}}=480$.

Also, the number of bacteria will present at the end of ${{n}^{th}}$ hour is given by ${{a}_{n+1}}=a{{r}^{n}}=\left( 30 \right){{2}^{n}}$.

31. What will Rs. $\mathbf{500}$ amounts to $\mathbf{10}$ years after its deposit in a bank which pays an annual interest rate of $\mathbf{10}%$ compounded annually?

Ans:

The deposited amount to the bank is Rs. $500$.

The amount obtained after the end of the first year 

$=$Rs. $500\left( 1+\dfrac{1}{10} \right)=Rs.\,500\left( 1.1 \right)$.

The amount obtained after the end of the second year,

$=Rs.\,500\left( 1.1 \right)\left( 1.1 \right)$.

The amount obtained after the end of third year,

$=Rs.\,500\left( 1.1 \right)\left( 1.1 \right)\left( 1.1 \right)$ and …so on

Therefore, after the end of $10$ years, the amount $=Rs.\,\,500\left( 1.1 \right)\left( 1.1 \right)...\left( 10\,\,\text{times} \right)$

$=Rs.\,\,500{{\left( 1.1 \right)}^{10}}$.

32. If A.M. and G.M. of roots of a quadratic equation are $\mathbf{8}$ and $\mathbf{5}$, respectively, then obtain the quadratic equation.

Ans:

First suppose that $a$ and $b$ are the roots of the quadratic equation.

Then, A.M. and G.M. of the roots $a$ and $b$ are such that

A.M.$=\dfrac{a+b}{2}$ and

G.M.$=\sqrt{ab}$.

Therefore, by the condition given to us,

$\dfrac{a+b}{2}=8$

$\Rightarrow a+b=16$                                                   …… (i)

and $\sqrt{ab}=5$

$\Rightarrow ab=25$                                                      …… (ii)

Now, we know that, in a quadratic equation of $x$,

${{x}^{2}}-x\left( \text{sum of roots} \right)+\left( \text{product of the roots} \right)=0$

$\Rightarrow {{x}^{2}}-\left( a+b \right)x+\left( ab \right)=0$

$\Rightarrow {{x}^{2}}-16x+25=0$, substituting the values of $a+b$ and $ab$.

Hence, the needed quadratic equation is ${{x}^{2}}-16x+25=0$.

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 9 Exercises

NCERT Solutions for Class 11 Math Chapter 9 Sequences and Series Exercise 9.3

Opting for the NCERT solutions for Ex 9.3 Class 11 Math is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.3 Class 11 Math NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Subject Sequences and Series textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Math Chapter 9 Exercise 9.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Math Chapter 9 Exercise 9.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 11 Math Chapter 9 Exercise 9.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is that they can be accessed both online and offline as well.

Topics Covered in Exercise 9.3 of Chapter 9 Sequences and Series 

Out of the four total exercises and a miscellaneous exercise, here are the NCERT Solutions for the third exercise of Class 11 Maths Chapter 9 Sequences and Series. Before solving them, you must know the topics covered in Exercise 9.3, which are given below:

• Geometric Progression (G.P)

• The general term of a G.P

• Sum to n terms of a G.P

• Geometric Mean (G.M)

• Relationship Between Arithmetic Mean (A.M) and G.M.

Important Formulas to Solve Class 11 Maths Chapter 9 Sequences and Series Exercise 9.3

If you’ve forgotten or need to brush up on the important formulas you will need to solve Exercise 9.3 of NCERT Maths Class 11 Chapter 9, we’ve got you covered. Revise them from the table below.