If S be the sum, P the product and R be the sum of the reciprocals of n terms of a GP, then P² is equal to
(a) S/R
(b) R/S
(c) (R/S)ⁿ
(d) (S/R)ⁿ

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Solution

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

$Sum of n terms of the G . P ., S = \frac{a (r^{n} - 1)}{(r - 1)} Product of n terms of the G . P ., P = a^{n} r^{[\frac{n (n - 1)}{2}]} Sum of the reciprocals of n terms of the G . P ., R = \frac{[\frac{1}{r^{n}} - 1]}{a (\frac{1}{r} - 1)} = \frac{(r^{n} - 1)}{a r^{(n - 1)} (r - 1)} ∴ P^{2} = {\{a^{2} r^{\frac{2 (n - 1)}{2}}\}}^{n} \Rightarrow P^{2} = {\{\frac{\frac{a (r^{n} - 1)}{(r - 1)}}{\frac{(r^{n} - 1)}{a r^{(n - 1)} (r - 1)}}\}}^{n} \Rightarrow P^{2} = {\{\frac{S}{R}\}}^{n}$

$Let the first term of the G . P . be a and the common ratio be r . Sum of n terms, S = \frac{a (r^{n} - 1)}{r - 1} Product of the G . P ., P = a^{n} r^{\frac{n (n + 1)}{2}} Sum of the reciprocals of n terms, R = \frac{(\frac{1}{r^{n} - 1})}{a (\frac{1}{r^{} - 1})} = \frac{(\frac{1 - r^{n}}{r^{n}})}{a (\frac{1 - r}{r})} p^{2} = {\{a^{2} r^{\frac{(n + 1)}{2}}\}}^{n} p^{2} = {\{\frac{\frac{a (r^{n} - 1)}{r - 1}}{\frac{(\frac{1 - r^{n}}{r^{n}})}{a (\frac{1 - r}{r})}}\}}^{n} = {\{\frac{S}{R}\}}^{n}$

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