$10 \cdot 27$ Let $S_{1}$ be the sum of the first $n$ terms of the arithmetic sequence $8,12, \cdots$, and $S_{2}$ be the sum of the first $n$ terms of the arithmetic sequence $17,19, \cdots$. If $n \neq 0$, then $S_{1}=S_{2}$
(A) is impossible for any $n$ value.
(B) is true for one $n$ value.
(C) is true for two $n$ values.
(D) is true for four $n$ values.
(E) is true for more than four $n$ values.
(17th American High School Mathematics Examination, 1966)
[Solution] From the given conditions and the formula for the sum of the first $n$ terms of an arithmetic sequence with the first term $a$ and common difference $d$, we have
$$
S_{n}=\frac{n}{2}[2 a+(n-1) d]
$$

Thus,
$$
\begin{array}{l}
S_{1}=\frac{n}{2}(12+4 n)=2 n^{2}+6 n, \\
S_{2}=\frac{n}{2}(32+2 n)=n^{2}+16 n .
\end{array}
$$

Therefore, from $S_{1}=S_{2}$, i.e., $2 n^{2}+6 n=n^{2}+16 n$, we have $n^{2}-10 n=0$. Since $n \neq 0$, we have $n-10=0$, which gives $n=10$. Hence, the answer is $(B)$.