2. 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 $S_{n}\left(n \in \mathbf{N}_{+}\right)$ is ( ).
(A) $S_{10}$
(B) $S_{11}$
(C) $S_{20}$
(D) $S_{21}$
2.C.

Let the common difference of the arithmetic sequence be $d$. According to the problem, we have $3\left(a_{1}+7 d\right)=5\left(a_{1}+12 d\right)$, solving this gives $d=-\frac{2}{39} a_{1}$. Thus, $a_{n}=a_{1}+(n-1) d=a_{1}-\frac{2}{39} a_{1}(n-1)$ $=a_{1}\left(\frac{41}{39}-\frac{2}{39} n\right)$.
To maximize $S_{n}\left(n \in \mathbf{N}_{+}\right)$, it is necessary that $a_{n} \geqslant 0$, which means $n \leqslant 20$.