1. If the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{1}{2}, a_{n+1}=\frac{2 a_{n}}{3 a_{n}+2}, n \in \mathbf{N}_{+}$, then $a_{2017}=$ $\qquad$ .
$$
\frac{1}{a_{n+1}}=\frac{1}{a_{n}}+\frac{3}{2} \Rightarrow \frac{1}{a_{n}}=\frac{3 n+1}{2} \Rightarrow a_{n}=\frac{2}{3 n+1} \Rightarrow a_{2017}=\frac{1}{3026} .
$$