NCERT Solutions for Class 11 Maths Chapter 8 - Sequences and Series Exercise 8.2

Exercise 8.2

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 ${{x}^{2}}-16x+25=0$.

Conclusion

In conclusion, Class 11 Maths Chapter 8 Exercise 8.2 Solutions has provided you with valuable practice in understanding and solving problems related to arithmetic progressions (AP). By working through Class 11 Maths NCERT Solutions Chapter 8 Exercise 8.2, you have learned how to identify arithmetic sequences, find the common difference, and calculate both the nth term and the sum of the first n terms of an AP. These concepts are important for your understanding of sequences and series, and they have various applications in real-world scenarios and advanced mathematical studies.

Class 11 Maths Chapter 8: Exercises Breakdown

CBSE Class 11 Maths Chapter 8 Other Study Materials

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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