2. Given that the first $n$ terms sum of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are $S_{n}$ and $T_{n}$ respectively, and for all positive integers $n$, $\frac{a_{n}}{b_{n}}=\frac{2 n-1}{3 n+1}$. Then $\frac{S_{6}}{T_{5}}=$ $\qquad$ .
2. $\frac{18}{25}$.

According to the problem, let
$$
a_{n}=k(2 n-1), b_{n}=k(3 n+1) \text {, }
$$

where, the constant $k \neq 0$.
Then $S_{6}=36 k, T_{5}=50 k$.
Thus, $\frac{S_{6}}{T_{5}}=\frac{36 k}{50 k}=\frac{18}{25}$.