Question

Solve the question given in attachment​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that the n terms of geometric progression be

$\red{\rm :\longmapsto\:a,ar, {ar}^{2}, - - - - - - , {ar}^{n - 1} } \:$

Now, we have

$\rm :\longmapsto\:S = a + ar + {ar}^{2} + {ar}^{3} + - - - + {ar}^{n - 1}$

We know, Sum of n terms of GP series having first term a and common ratio r respectively is

$\boxed{ \tt{ \: S_n \: = \: \frac{a( {r}^{n} - 1) }{r - 1} \: }}$

So, using this identity, we get

$\rm \implies\:\boxed{ \tt{ \: S = \frac{a( {r}^{n} - 1)}{r - 1} \: }} - - - - (1)$

Now, we have

$\rm :\longmapsto\:R = \dfrac{1}{a} + \dfrac{1}{ar} + \dfrac{1}{ {ar}^{2} } + - - - + \dfrac{1}{ {ar}^{n - 1} }$

$\rm :\longmapsto\:R = \dfrac{1}{a} \times \dfrac{1 - {\bigg[\dfrac{1}{r} \bigg]}^{n} }{1 - \dfrac{1}{r} }$

$\rm :\longmapsto\:R = \dfrac{1}{a} \times \dfrac{1 - \dfrac{1}{ {r}^{n} } }{ \dfrac{r - 1}{r} }$

$\rm :\longmapsto\:R = \dfrac{1}{a} \times \dfrac{\dfrac{ {r}^{n} - 1}{ {r}^{n} } }{ \dfrac{r - 1}{r} }$

$\rm :\longmapsto\:R = \dfrac{1}{a} \times \dfrac{ {r}^{n} - 1}{ {r}^{n} } \times \dfrac{r}{r - 1}$

$\rm :\longmapsto\:R = \dfrac{1}{a} \times \dfrac{ {r}^{n} - 1}{ {r}^{n - 1} } \times \dfrac{1}{r - 1}$

$\rm \implies\:\boxed{ \tt{ \: R = \dfrac{ {r}^{n} - 1}{ a \: {r}^{n - 1} \: (r - 1) } \: }}$

Now, we have

$\rm :\longmapsto\:P = a.ar. {ar}^{2}. {ar}^{3}... {ar}^{n - 1}$

$\rm :\longmapsto\:P = {a}^{n}. {r}^{1 + 2 + 3 + - - - + (n - 1)}$

$\rm :\longmapsto\:P = {a}^{n}. {\bigg(r\bigg) }^{\dfrac{n - 1}{2} (1 + n - 1)}$

$\rm :\longmapsto\:P = {a}^{n}. {\bigg(r\bigg) }^{\dfrac{n(n - 1)}{2}}$

$\rm \implies\:\boxed{ \tt{ \: {P}^{2} = {a}^{2n}. {r}^{n(n - 1)} \: }} - - - - (3)$

Now, Consider,

$\rm :\longmapsto\: {\bigg[\dfrac{S}{R} \bigg]}^{n}$

$\rm \:  =  \: {\bigg[\dfrac{a( {r}^{n} - 1)}{r - 1} \times \dfrac{ {ar}^{n - 1}(r - 1) }{ {r}^{n} - 1 } \bigg]}^{n}$

$\rm \:  =  \: {\bigg[ {a}^{2} {r}^{n - 1} \bigg]}^{n}$

$\rm \:  =  \: {a}^{2n}. {r}^{n(n - 1)}$

$\rm \:  =  \: {P}^{2}$

Thus,

$\rm :\longmapsto\: \boxed{ \tt{ \: {\bigg[\dfrac{S}{R} \bigg]}^{n} = {P}^{2} \: }}$

Hence, Proved