If S is the sum,P is the product,and R is the | vedclass

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(A) Let the $n$ terms of the geometric progression be $a, ar, ar^2, \dots, ar^{n-1}$.$S = a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$

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

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(A) Let the $n$ terms of the geometric progression be $a, ar, ar^2, \dots, ar^{n-1}$.$S = a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$.$P = a \cdot (ar) \cdot (ar^2) \cdot \dots \cdot (ar^{n-1}) = a^n r^{0+1+2+\dots+(n-1)} = a^n r^{\frac{n(n-1)}{2}}$.$P^2 = a^{2n} r^{n(n-1)}$.$R = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \dots + \frac{1}{ar^{n-1}} = \frac{r^{n-1} + r^{n-2} + \dots + 1}{ar^{n-1}} = \frac{1}{ar^{n-1}} \cdot \frac{r^n - 1}{r - 1}$.Now,$\frac{S}{R} = \frac{a(r^n - 1)}{r - 1} \cdot \frac{ar^{n-1}(r - 1)}{r^n - 1} = a^2 r^{n-1}$.Therefore,$(\frac{S}{R})^n = (a^2 r^{n-1})^n = a^{2n} r^{n(n-1)}$.Comparing this with $P^2$,we get $P^2 = (\frac{S}{R})^n$.

If $S$ is the sum,$P$ is the product,and $R$ is the sum of the reciprocals of $n$ terms of a geometric progression,then $P^2 = \dots$

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