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 $S = a + ar + \dots + ar^{n-1} = \frac{a(1-r^n)}{1-r}$.The product $P =

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

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(D) Let the $n$ terms of the $G.P.$ be $a, ar, ar^2, \dots, ar^{n-1}$.The sum $S = a + ar + \dots + ar^{n-1} = \frac{a(1-r^n)}{1-r}$.The product $P = a \cdot ar \cdot ar^2 \dots ar^{n-1} = a^n r^{0+1+2+\dots+(n-1)} = a^n r^{\frac{n(n-1)}{2}}$.Thus,$P^2 = a^{2n} r^{n(n-1)}$.The sum of reciprocals $R = \frac{1}{a} + \frac{1}{ar} + \dots + \frac{1}{ar^{n-1}} = \frac{1}{a} \left( 1 + \frac{1}{r} + \dots + \frac{1}{r^{n-1}} \right) = \frac{1}{a} \left( \frac{1 - (1/r)^n}{1 - 1/r} \right) = \frac{1}{a} \left( \frac{(r^n-1)/r^n}{(r-1)/r} \right) = \frac{r^n-1}{a r^{n-1} (r-1)} = \frac{1-r^n}{a r^{n-1} (1-r)}$.Now,$\frac{S}{R} = \frac{a(1-r^n)}{1-r} \div \frac{1-r^n}{a r^{n-1} (1-r)} = a^2 r^{n-1}$.Therefore,$(\frac{S}{R})^n = (a^2 r^{n-1})^n = a^{2n} r^{n(n-1)} = P^2$.

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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