If the sum of the n terms of a G.P. is S,the | vedclass

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(D) Let the $n$ terms of the $G.P.$ be $a, ar, ar^2, \dots, ar^{n-1}$.The sum $S$ is given by $S = \frac{a(1 - r^n)}{1 - r}$ ......$(i)$The product $P

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${\left( {\frac{S}{R}} \right)^n}$

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(D) Let the $n$ terms of the $G.P.$ be $a, ar, ar^2, \dots, ar^{n-1}$.The sum $S$ is given by $S = \frac{a(1 - r^n)}{1 - r}$ ......$(i)$The product $P$ is given by $P = a \cdot (ar) \cdot (ar^2) \dots (ar^{n-1}) = a^n r^{0+1+2+\dots+(n-1)} = a^n r^{\frac{n(n-1)}{2}}$.Squaring both sides,we get $P^2 = a^{2n} r^{n(n-1)}$ ......(ii)The sum of the reciprocals $R$ is given by $R = \frac{1}{a} + \frac{1}{ar} + \dots + \frac{1}{ar^{n-1}}$.This is a $G.P.$ with first term $\frac{1}{a}$ and common ratio $\frac{1}{r}$.$R = \frac{\frac{1}{a}(1 - (\frac{1}{r})^n)}{1 - \frac{1}{r}} = \frac{\frac{1}{a}(\frac{r^n - 1}{r^n})}{\frac{r - 1}{r}} = \frac{r^n - 1}{a r^n} \cdot \frac{r}{r - 1} = \frac{r^n - 1}{a r^{n-1}(r - 1)} = \frac{1 - r^n}{a r^{n-1}(1 - r)}$ ......(iii)Now,$\frac{S}{R} = \frac{a(1 - r^n)}{1 - r} \cdot \frac{a r^{n-1}(1 - r)}{1 - r^n} = a^2 r^{n-1}$.Therefore,$(\frac{S}{R})^n = (a^2 r^{n-1})^n = a^{2n} r^{n(n-1)}$.Comparing this with equation (ii),we get $P^2 = (\frac{S}{R})^n$.

If the sum of the $n$ terms of a $G.P.$ is $S$,the product is $P$,and the sum of their reciprocals is $R$,then ${P^2}$ is equal to:

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