8. In the arithmetic sequence $\left\{a_{n}\right\}$, for any positive integer $n$, it holds that $a_{n}+2 a_{n+1}+3 a_{n+2}=6 n+22$, then $a_{2017}=$ $\qquad$
Let $a_{n}=k n+b$, then $k n+b+2 k(n+1)+2 b+3 k(n+2)+3 b=6 k n+8 k+6 b$ $=6 n+22 \Rightarrow k=1, b=\frac{7}{3}$. Therefore, $a_{2017}=2017+\frac{7}{3}=\frac{6058}{3}$.