$10 \cdot 48$ Suppose there is an $n$-term geometric sequence, the first term is 1, the common ratio is $r$, and the sum is $s$, where $r$ and $s$ are non-zero numbers. The sum of the new geometric sequence obtained by replacing each term of the original sequence with its reciprocal is
(A) $\frac{1}{s}$.
(B) $\frac{1}{r^{n} s}$.
(C) $\frac{s}{r^{n-1}}$.
(D) $\frac{r^{n}}{s}$.
(E) $\frac{r^{n-1}}{s}$.
(27th American High School Mathematics Examination, 1976)
[Solution] The sum of the terms in the new sequence is
$$
1+\frac{1}{r}+\cdots+\frac{1}{r^{n-1}}=\frac{r^{n-1}+r^{n-2}+\cdots+1}{r^{n-1}}=\frac{s}{r^{n-1}} \text {. }
$$

Therefore, the answer is $(C)$.