Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms in a $G.P.$ Prove that $P ^{2} R ^{n}= S ^{n}$

Let the $G.P.$ be $a, a r, a r^{2}, a r^{3} \ldots . . a r^{n-1}$

According to the given information,

$S=\frac{a\left(r^{n}-1\right)}{r-1}$

$P=a^{n} \times r^{1+2+\ldots+n-1}$

$=a^{n} r^{\frac{n(n-1)}{2}}$               [ $\because $ Sum of first $4n$ natural number is $n \frac{(n+1)}{2}$ ]

$R=\frac{1}{a}+\frac{1}{a r}+\ldots \ldots+\frac{1}{a r^{n-1}}$

$=\frac{r^{n-1}+r^{n-2}+\ldots . r+1}{a r^{n-1}}$

$=\frac{1\left(r^{n}-1\right)}{(r-1)} \times \frac{1}{a r^{n-1}}$    [ $\because $ $1, r, \ldots \ldots r^{n-1}$ forms a $G.P.$ ]

$=\frac{r^{n}-1}{a r^{n-1}(r-1)}$

$\therefore P^{2} R^{n}=a^{2 n} r^{n(n-1)} \frac{\left(r^{n}-1\right)^{n}}{a^{n} r^{n(n-1)}(r-1)^{n}}$

$=\frac{a^{n}\left(r^{n}-1\right)^{n}}{(r-1)^{n}}$

$=\left[\frac{a\left(r^{n}-1\right)}{(r-1)}\right]^{n}$

$=S^{n}$

Hence, $P^{2} R^{n}=S^{n}$