smc

Let the geometric sequence have first term $a$ and common ratio $R$. Then, the first $n$ terms of the sequence are $a,aR,a{R}^2,\dots ,a{R}^{n-1}$. The product of these terms $P$ is $a^n{R}^{1+2+\dots +n-1}=a^n{R}^{\frac{(n-1)n}{2}}$ by the formula for triangular numbers. Using the sum formula reveals that $S=a\cdot \frac{1-R^n}{1-R}$. We know that $S^{\prime }=\frac{1}{a}+\frac{1}{aR}+\dots +\frac{1}{a{R}^{n-1}}$ Combining fractions reveals that $S^{\prime }=\frac{a{R}^{n-1}+a{R}^{n-2}+\dots +aR+a}{a^2{R}^{n-1}}=\frac{S}{a^2{R}^{n-1}}$. Note that this denominator looks suspiciously similar to our formula for $P$. In fact, $(a^2+R^{n-1}{)}^{\frac{n}{2}}=a^n+R^{\frac{(n-1)n}{2}}=P$. Because $(S/S^{\prime }{)}^{\frac{n}{2}}=(a^2+R^{n-1}{)}^{\frac{n}{2}}=P$, our answer is $\boxed{(S/S^{\prime }{)}^{\frac{1}{2}n}}$.