1. Let the sequence $\left\{a_{n}\right\}$ satisfy:
$$
a_{1}=\frac{1}{2}, a_{n+1}=\frac{2 a_{n}}{3 a_{n}+2}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Then the value of $a_{2017}$ is $\qquad$ .
$-1 . \frac{1}{3026}$.
From $a_{n+1}=\frac{2 a_{n}}{3 a_{n}+2} \Rightarrow \frac{1}{a_{n+1}}=\frac{1}{a_{n}}+\frac{3}{2}$.
Also, $a_{1}=\frac{1}{2}$, so
$$
\begin{array}{l}
\frac{1}{a_{2017}}=2+2016 \times \frac{3}{2}=3026 \\
\Rightarrow a_{2017}=\frac{1}{3026} .
\end{array}
$$