If S be the sum, P the product and R be the sum of the reciprocals of n terms of a GP,

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

Sum of $n$ terms of the G.P., $\mathrm{S}=\frac{a\left(r^{n}-1\right)}{(r-1)}$

Product of $n$ terms of the G.P., $\mathrm{P}=a^{n} r\left[\frac{n(n-1)}{2}\right]$

Sum of the reciprocals of $n$ terms of the G.P., $\mathrm{R}=\frac{\left[\frac{1}{r^{n}}-1\right]}{a\left(\frac{1}{r}-1\right)}=\frac{\left(r^{n}-1\right)}{a r^{(n-1)}(r-1)}$

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

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

$\Rightarrow \mathrm{P}^{2}=\left\{\frac{\mathrm{S}}{\mathrm{R}}\right\}^{\mathrm{n}}$

Let the first term of the G.P. be $a$ and the common ratio be $r$.

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

Product of the G.P., $\mathrm{P}=a^{n} r^{\frac{n(n+1)}{2}}$

Sum of the reciprocals of $n$ terms, $R=\frac{\left(1 / \mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{a}(1 / \mathrm{r}-1)}=\frac{\left(\frac{1-\mathrm{r}^{\mathrm{n}}}{\mathrm{r}^{\mathrm{n}}}\right)}{\mathrm{a}\left(\frac{1-\mathrm{r}}{\mathrm{r}}\right)}$

$\mathrm{p}^{2}=\left\{\mathrm{a}^{2} \mathrm{r}^{\frac{(\mathrm{n}+1)}{2}}\right\}^{\mathrm{n}}$

$\mathrm{p}^{2}=\left\{\frac{\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}-1}\right.}{\mathrm{r}-1}}{\frac{\left(\frac{1-\mathrm{n}^{\mathrm{n}}}{\mathrm{r}^{\mathrm{n}}}\right)}{\mathrm{a}\left(\frac{1-\mathrm{r}}{\mathrm{r}}\right)}}\right\}=\left\{\frac{\mathrm{S}}{\mathrm{R}}\right\}^{\mathrm{n}}$