NCERT Solutions for Class 11 Maths Chapter 8 Sequences And Series Ex 8.2

Exercise 8.2

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

Ans:

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

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

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

$\begin{array}{rl}& a_{20}=a{r}^{20-1}\\ & ={\displaystyle \frac{5}{2}}{\left({\displaystyle \frac{1}{2}}\right)}^{19}\\ & ={\displaystyle \frac{5}{2\times 2^{19}}}\\ & ={\displaystyle \frac{5}{2^{20}}}.\end{array}$ 

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

$\begin{array}{rl}& a_n=a{r}^{n-1}\\ & ={\displaystyle \frac{5}{2}}{\left({\displaystyle \frac{1}{2}}\right)}^{n-1}\\ & ={\displaystyle \frac{5}{2\cdot 2^{n-1}}}\\ & ={\displaystyle \frac{5}{2^n}}.\end{array}$

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

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$

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

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

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

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

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

$\begin{array}{rl}& a_{12}=a{r}^{12-1}\\ & =a{r}^{11}\\ & ={\displaystyle \frac{3}{2}}{\left(2\right)}^{11}\\ & =3{\left(2\right)}^{10}\\ & =3072.\end{array}$

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 $\mathbf{p}\mathbf{,}\mathbf{q}\mathbf{,}$ and $\mathbf{s}$ respectively. Show that ${\mathbf{q}}^{\mathbf{2}}\mathbf{=}\mathbf{ps}$.

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

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

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

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

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

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

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

${\displaystyle \frac{q}{p}}={\displaystyle \frac{s}{q}}$

$\Rightarrow q^2=ps$.

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 $\mathbf{-}\mathbf{3}$. 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,

$a_4=a{r}^3=(-3)r^3$ and $a_2=a{r}^1=(-3)r$.

So, by the condition provided to us,

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

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

$\Rightarrow r=-3$.

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

$\begin{array}{rl}& a_7=(-3){(-3)}^{7-1}\\ & ={(-3)}^7\\ & =-2187.\end{array}$

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

5. Which term of the following sequences:

(a) $\mathbf{2}\mathbf{,}\mathbf{2}\sqrt{\mathbf{2}}\mathbf{,}\mathbf{4}\mathbf{,}...$ 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 $r={\displaystyle \frac{2\sqrt{2}}{2}}=\sqrt{2}$.

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

Therefore, we have

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

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

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

$\Rightarrow {\displaystyle \frac{n+1}{2}}=6$

$\Rightarrow n+1=12$

$\Rightarrow n=11$.

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

(b) $\sqrt{\mathbf{3}}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{3}\sqrt{\mathbf{3}}\mathbf{,}...$ 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 $r={\displaystyle \frac{3}{\sqrt{3}}}=\sqrt{3}$.

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

Therefore, we have

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

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

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

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

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

$\Rightarrow n=12$.

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

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

Ans:

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

$r={\displaystyle \frac{{\displaystyle \frac{1}{9}}}{{\displaystyle \frac{1}{3}}}}={\displaystyle \frac{1}{3}}$.

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

Therefore, we have

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

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

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

$\Rightarrow n=9$.

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

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

Ans:

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

Now, let $a_1=-{\displaystyle \frac{2}{7}}$, $a_2=x$ and $a_3=-{\displaystyle \frac{7}{2}}$.

Then, ${\displaystyle \frac{a_2}{a_1}}={\displaystyle \frac{x}{-{\displaystyle \frac{2}{7}}}}={\displaystyle \frac{-7x}{2}}$ and ${\displaystyle \frac{a_3}{a_2}}={\displaystyle \frac{-{\displaystyle \frac{7}{2}}}{x}}=-{\displaystyle \frac{7}{2x}}$.

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

$r={\displaystyle \frac{a_2}{a_1}}={\displaystyle \frac{a_3}{a_2}}$

$\Rightarrow {\displaystyle \frac{-7x}{2}}=-{\displaystyle \frac{7}{2x}}$

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

$\Rightarrow x=\sqrt{1}$

$\Rightarrow x=\pm 1$.

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

7. Find the sum of the numbers $\mathbf{0}\mathbf{.15}\mathbf{,}\mathbf{0}\mathbf{.015}\mathbf{,}\mathbf{0}\mathbf{.0015}\mathbf{,}...$ 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 $r={\displaystyle \frac{0.015}{0.15}}=0.1$

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

$S_n={\displaystyle \frac{a(1-r^n)}{1-r}}$

Therefore,

$S_{20}={\displaystyle \frac{0.15[1-{\left(0.1\right)}^{20}]}{1-0.1}}$

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

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

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

8. Find the sum of the numbers $\sqrt{\mathbf{7}}\mathbf{,}\sqrt{\mathbf{21}}\mathbf{,}\mathbf{3}\sqrt{\mathbf{7}}\mathbf{,}...$ 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 $r={\displaystyle \frac{\sqrt{21}}{\sqrt{7}}}=\sqrt{3}$.

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

$S_n={\displaystyle \frac{a(1-r^n)}{1-r}}$

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

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

$S_n={\displaystyle \frac{\sqrt{7}(1+\sqrt{3})}{2}}[{\left(3\right)}^{{\displaystyle \frac{n}{2}}}-1]$.

9. Find the sum of the numbers $\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{a}\mathbf{,}{\mathbf{a}}^{\mathbf{2}}\mathbf{,}\mathbf{-}{\mathbf{a}}^{\mathbf{3}}\mathbf{,}...$ (if $\mathbf{a}\ne \mathbf{-}\mathbf{1}$) 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 $r={\displaystyle \frac{-a}{1}}=-a$.

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

$S_n={\displaystyle \frac{a(1-r^n)}{1-r}}$

$={\displaystyle \frac{1[1-{(-a)}^n]}{1-(-a)}}$

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

$S_n={\displaystyle \frac{[1-{(-a)}^n]}{1+a}}$.

10. Find the sum of the numbers ${\mathbf{x}}^{\mathbf{3}}\mathbf{,}{\mathbf{x}}^{\mathbf{5}}\mathbf{,}{\mathbf{x}}^{\mathbf{7}}\mathbf{,}...$(if $\mathbf{x}\ne \pm \mathbf{1}$) 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 $r={\displaystyle \frac{x^5}{x^3}}=x^2$.

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

$S_n={\displaystyle \frac{a(1-r^n)}{1-r}}$

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

$={\displaystyle \frac{x^3(1-x^{2n})}{1-x^2}}$.

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  $S_n={\displaystyle \frac{x^3(1-x^{2n})}{1-x^2}}$.

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

Ans:

The given sum can be written as

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

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

Therefore, we have

$S_n={\displaystyle \frac{a(r^n-1)}{r-1}}$

$\Rightarrow S_n={\displaystyle \frac{3[3^{11}-1]}{3-1}}$

$\Rightarrow S_n={\displaystyle \frac{3}{2}}(3^{11}-1)$

Thus,

$\sum _{k=1}^{11}{3}^k={\displaystyle \frac{3}{2}}(3^{11}-1)$.                                                 …… (ii)

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

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

12. The sum of the first three terms of a G.P. is ${\displaystyle \frac{\mathbf{39}}{\mathbf{10}}}$ 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 ${\displaystyle \frac{a}{r}},a,ar$.

Now, by the given conditions,

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

$\left({\displaystyle \frac{a}{r}}\right)\cdot \left(a\right)\cdot \left(ar\right)=1$                                    …… (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

${\displaystyle \frac{1}{r}}+1+r={\displaystyle \frac{39}{10}}$

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

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

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

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

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

$\Rightarrow r={\displaystyle \frac{2}{5}}$ or ${\displaystyle \frac{5}{2}}$.

Hence, ${\displaystyle \frac{5}{2}},$ $1$, and ${\displaystyle \frac{2}{5}}$ 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.,

$S_n={\displaystyle \frac{a(r^n-1)}{r-1}}$.

Therefore,

$\begin{array}{rl}& {\displaystyle \frac{3(3^n-1)}{3-1}}=120\\ & \Rightarrow 3^n-1=120\times {\displaystyle \frac{2}{3}}\\ & \Rightarrow 3^n-1=80\\ & \Rightarrow 3^n=81\\ & \Rightarrow 3^n=3^4\end{array}$

$\Rightarrow n=4$.

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,

$a+ar+a{r}^2=16$              

 $\Rightarrow a(1+r+r^3)=16$                                           …… (i)

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

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

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

${\displaystyle \frac{a{r}^3(1+r+r^2)}{a(1+r+r^2)}}={\displaystyle \frac{128}{16}}$

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

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

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

$a(1+2+2^2)=16$

$\begin{array}{rl}& \Rightarrow a\times 7=16\\ & \Rightarrow a={\displaystyle \frac{16}{7}}\end{array}$

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

$S_n={\displaystyle \frac{a(r^n-1)}{r-1}}$

$={\displaystyle \frac{16}{7}}{\displaystyle \frac{(2^n-1)}{2-1}}$

$={\displaystyle \frac{16}{7}}(2^n-1)$.

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, $a_7=\left(729\right)r^{7-1}=729\cdot r^6$

$\begin{array}{rl}& \Rightarrow 64=729r^6\\ & \Rightarrow r^6={\left({\displaystyle \frac{2}{3}}\right)}^6\\ & \Rightarrow r={\displaystyle \frac{2}{3}}.\end{array}$

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

$S_n={\displaystyle \frac{a(1-r^n)}{1-r}}$

Therefore, for $n=7$, we have

$S_7={\displaystyle \frac{729[1-{\left({\displaystyle \frac{2}{3}}\right)}^7]}{1-{\displaystyle \frac{2}{3}}}}$

$=3^7\times {\displaystyle \frac{3^7-2^7}{3^7}}$

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

$=2059$.

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,

$-4={\displaystyle \frac{a(1-r^2)}{1-r}}$                                                    …… (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

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

$\Rightarrow -4={\displaystyle \frac{a(1-4)}{-1}}$

$\Rightarrow -4=3a$

$\Rightarrow a=-{\displaystyle \frac{4}{3}}$.

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

$-4={\displaystyle \frac{a[1-{(-2)}^2]}{1-(-2)}}$

$\Rightarrow -4={\displaystyle \frac{a(1-4)}{1+2}}$

$\Rightarrow -4={\displaystyle \frac{a(-3)}{3}}$

$\Rightarrow a=4$.

Hence, the G.P. can be of the form $-{\displaystyle \frac{4}{3}},-{\displaystyle \frac{8}{3}},-{\displaystyle \frac{16}{3}},...$ 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 $\mathbf{x}\mathbf{,}\mathbf{y}$ 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

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

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

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

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

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

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

${\displaystyle \frac{y}{x}}={\displaystyle \frac{z}{y}}$.

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

$S_n=8+88+888+8888+\cdot \cdot \cdot \text{to n terms}$

$={\displaystyle \frac{8}{9}}[9+99+999+9999+\cdot \cdot \cdot \text{to n terms}]$

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

$={\displaystyle \frac{8}{9}}[(10+{10}^2+\cdot \cdot \cdot +\text{n terms})-(1+1+1+\cdot \cdot \cdot \text{n terms})]$

$\begin{array}{rl}& ={\displaystyle \frac{8}{9}}[{\displaystyle \frac{10({10}^n-1)}{10-1}}-n]\\ & ={\displaystyle \frac{80}{81}}({10}^n-1)-{\displaystyle \frac{8}{9}}n.\end{array}$

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

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 $\mathbf{128},\mathbf{32},\mathbf{8},\mathbf{2},{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}$.

Ans:

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

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

$=64[4+2+1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2^2}}]$.

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

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

$S_n={\displaystyle \frac{a(1-r^n)}{1-r}}$, $r<1$.

Therefore,

$\begin{array}{rl}& S_5={\displaystyle \frac{4[1-{\left({\displaystyle \frac{1}{2}}\right)}^5]}{1-{\displaystyle \frac{1}{2}}}}\\ & ={\displaystyle \frac{4[1-{\displaystyle \frac{1}{32}}]}{{\displaystyle \frac{1}{2}}}}\\ & =8\left({\displaystyle \frac{32-1}{32}}\right)\\ & ={\displaystyle \frac{31}{4}}.\end{array}$

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

20. Show that the products of the corresponding terms of the sequences $\mathbf{a}\mathbf{,}\mathbf{ar}\mathbf{,}\mathbf{a}{\mathbf{r}}^{\mathbf{2}}\mathbf{,}...\mathbf{,}\mathbf{a}{\mathbf{r}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$ 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. ${\displaystyle \frac{\text{Second term }}{\text{First term}}}={\displaystyle \frac{arAR}{aA}}=rR$                                  …… (i)

Also,

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

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

${\displaystyle \frac{\text{Second term }}{\text{First term}}}={\displaystyle \frac{\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

$a(r^2-1)=9$                                                …… (iii)

$ar(1-r^2)=18$                                              …… (iv)

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

${\displaystyle \frac{ar(1-r^2)}{-a(1-r^2)}}={\displaystyle \frac{18}{9}}$

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

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

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

Hence, the first four numbers that forms a geometric progression are $3,3(-2),3{(-2)}^2,$ and $3{(-2)}^3$, 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 ${\mathbf{a}}^{\mathbf{q}\mathbf{-}\mathbf{r}}{\mathbf{b}}^{\mathbf{r}\mathbf{-}\mathbf{p}}{\mathbf{c}}^{\mathbf{p}\mathbf{-}\mathbf{q}}\mathbf{=}\mathbf{1}$.

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

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

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

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

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,