［Solution］Let the sequence be $a, a r, a r^{2}, \cdots, a r^{n-1}$, then
$$
\begin{array}{c}
P=a^{n} r^{\frac{1}{2}(n-1) n}, \\
S=a \frac{1-r^{n}}{1-r}, \\
S^{\prime}=\frac{1}{a} \cdot \frac{1-r^{-n}}{1-r^{-1}}=\frac{r^{-(n-1)}}{a} \cdot \frac{1-r^{n}}{1-r} . \\
\therefore \quad \frac{S}{S^{\prime}}=a^{2} r^{n-1}, \quad \therefore \quad\left(\frac{S}{S^{\prime}}\right)^{\frac{1}{2} n}=a^{n} r^{\frac{1}{2}(n-1) n}=P .
\end{array}
$$

Therefore, the correct choice is $(B)$.
$10 \cdot 50$ If $n$ quantities form a geometric sequence, $P$ is their product, $S$ is their sum, and $S^{\prime}$ is the sum of their reciprocals, then $P$ in terms of $S, S^{\prime}$, and $n$ is
(A) $\left(S S^{\prime}\right)^{\frac{1}{2} n}$.
(B) $\left(\frac{S}{S^{\prime}}\right)^{\frac{1}{2} n}$.
(C) $\left(S S^{\prime}\right)^{n-2}$.
(D) $\left(\frac{S}{S^{\prime}}\right)^{n}$.
(E) $\left(\frac{S^{\prime}}{S}\right)^{\frac{1}{2}(n-1)}$.
(22nd American High School Mathematics Examination, 1971)