$13 \cdot 66$ Let the arithmetic sequence $\left\{a_{n}\right\}$ satisfy $3 a_{8}=5 a_{13}$ and $a_{1}>0, S_{n}$ be the sum of its first $n$ terms. Then the largest $S_{n}(n \in N)$ is
(A) $\mathrm{S}_{10}$.
(B) $S_{11}$.
(C) $\mathrm{S}_{20}$.
(D) $\mathrm{S}_{21}$.
(China High School Mathematics League, 1995)
［Solution］Let the common difference of the arithmetic sequence be $d$.
Then from $3 a_{8}=5 a_{13}$ we get $3\left(a_{1}+7 d\right)=5\left(a_{1}+12 d\right)$, which means $a_{1}=-19.5 d$.

Thus, $a_{20}=a_{1}+19 d>0$, and $a_{21}=a_{1}+20 d<0$. Therefore, the sum of the first 20 terms of the arithmetic sequence is the maximum.
Hence, the answer is $(C)$.