1. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{n+1} \leq \frac{a_{n+2}+a_{n}}{2}, a_{1}=1, a_{404}=2016$, then the maximum value of $a_{6}$ is $\qquad$
1. 26

Analysis: Let $b_{n}=a_{n+1}-a_{n}$, then $a_{n}=b_{n-1}+b_{n-1}+\mathrm{L}+b_{1}+a_{1}$, since $a_{n+1}-a_{n} \leq a_{n+2}-a_{n+1}$, hence $b_{n}$ is monotonically increasing;
$$
\begin{array}{l}
2016=a_{404}=b_{403}+b_{402}+\mathrm{L}+b_{1}+1, \text { so } b_{403}+b_{402}+\mathrm{L}+b_{1}=2015 ; \\
\text { Also, } b_{403}+b_{402}+\mathrm{L}+b_{1} \geq \frac{398}{5}\left(b_{5}+b_{4}+\mathrm{L}+b_{1}\right)+b_{5}+b_{4}+\mathrm{L}+b_{1}=\frac{403}{5}\left(b_{5}+b_{4}+\mathrm{L}+b_{1}\right),
\end{array}
$$
so $b_{5}+b_{4}+\mathrm{L}+b_{1} \leq 25$, that is, $a_{6}=b_{5}+b_{4}+\mathrm{L}+b_{1}+1 \leq 26, a_{6}=26$ if and only if $a_{n}=5 n-4$.