1. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=32, a_{n+1}-a_{n}=2 n\left(n \in \mathbf{Z}_{+}\right)$. Then the minimum value of $\frac{a_{n}}{n}$ is $\qquad$.
$-1 . \frac{31}{3}$.
From the given, we know
$$
\begin{array}{l}
a_{n}=a_{1}+\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right) \\
=32+\sum_{k=1}^{n-1} 2 k=n(n-1)+32 . \\
\text { Therefore, } \frac{a_{n}}{n}=n-1+\frac{32}{n} .
\end{array}
$$

When $n=5$, $\frac{a_{n}}{n}=\frac{52}{5}$;
When $n=6$, $\frac{a_{n}}{n}=\frac{31}{3}$.
In summary, when $n=6$, $\frac{a_{n}}{n}$ reaches its minimum value $\frac{31}{3}$.