If the sum of the n terms of G.P. is S product is P and sum of their inverse is R, than $P^{2}$ is equal to

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Solution

The correct option is D

$S = a \frac{(1 - r^{n})}{1 - r} P = a^{n} r \frac{n (n - 1)}{2} R = \frac{1}{a} + \frac{1}{a r} + \frac{1}{a r^{2}} + \dots \dots + \frac{1}{a r^{n - 1}} = \frac{\frac{1}{a} (1 - \frac{1}{r^{n}})}{1 - \frac{1}{r}} = \frac{1}{a r^{n - 1}} (\frac{r^{n} - 1}{r - 1}) p^{2} = a^{2 n} r^{n (n - 1)} (\frac{S}{R}) = a^{2 n} r^{n (n - 1)}$

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