Let the sum of $n$ terms be $a+a r+a r^2+$ where $r < 1$
Then $S=\frac{a\left(1-r^n\right)}{1-r} \quad \ldots(i)$
Also, $R=\frac{1}{a}+\frac{1}{a r}+\frac{1}{a r^2}+$.............to $n$ terms
$=\frac{\frac{1}{a}\left[\left(\frac{1}{r}\right)^n-1\right]}{\frac{1}{r}-1}\left[\because \frac{1}{r}>1\right.$ as $\left.r < 1\right]$
$=\frac{1}{a} \cdot \frac{1-r^n}{r^n} \cdot \frac{r}{1-r}=\frac{1-r^n}{a r^{(n-1)}(1-r)} \quad \ldots(ii)$
and
$\begin{aligned} P &=(a)(a r)\left(a r^2\right) \ldots \ldots \ldots \ldots\left(a r^{n-1}\right) \\=& a^n r^{1+2+3+\ldots \ldots \ldots \ldots \ldots(n-1)} \\ &=a^n r^{\frac{n-1}{2}}\left(\frac{2+(n-2)}{2}\right) \\ &=a^n \cdot r^{\frac{n-1}{2}(1+\overline{n-1})} \\ P &=a^n \cdot r^{n(n-1) / 2} \quad \ldots(iii) \end{aligned}$
L.H.S. : $P^2 R^n=a^{2 n} r^{n(n-1)} \cdot \frac{\left(1-r^n\right)^n}{a r^{n(n-1)}(1-r)^n}$
$=\frac{a^n\left(1-r^n\right)^n}{(1-r)^n}=\left[\frac{a\left(1-r^n\right)}{1-r}\right]^n$
$P^2 R^n=S^n$