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

A) $\left( \frac{S}{R} \right)^{n}$
B) $\frac{S}{R}$
C) $\left( \frac{R}{S} \right)^{n}$
D) $\frac{R}{S}$
Explanation

(a)
Let $S = a + ar + ar^{2} + ... + ar^{n - 1}$
$= \frac{a(1 - r^{n})}{1 - r}\ \ \ \ \ \lbrack\therefore r < 1\rbrack$
$P = a.ar.ar^{2}\ldots.ar^{n - 1} = a^{n}r^{1 + 2 + ... + (n - 1)}$
$= a^{n}r^{\frac{n(n - 1)}{2}}$
and $R = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^{2}} + \ldots + \frac{1}{ar^{n - 1}}$
$= \frac{\frac{1}{a}\left( \frac{1}{r^{n}} - 1 \right)}{\frac{1}{r} - 1} = \frac{\frac{1}{a}(1 - r^{n})}{r^{n - 1}(1 - r)}\ \ \ \left\lbrack \therefore\frac{1}{r} > 1 \right\rbrack$
Now, $\left( \frac{S}{R} \right)^{n} = \left\{ \frac{\frac{a\left( 1 - r^{n} \right)}{(1 - r)}}{\frac{\frac{1}{a}\left( 1 - r^{n} \right)}{r^{n - 1}(1 - r)}} \right\}^{n} = a^{2n}r^{n(n - 1)}$
$= \left\{ a^{n}r^{\frac{n(n - 1)}{2}} \right\}^{2} = P^{2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.