The JE integer sequence $a_{1}, a_{2}, \cdots$, satisfies $a:=120, a_{n+2}=a_{n}+a_{n-1}(n \geqslant 1)$, then the value of $a_{8}$ is
9. 194.

Given $a_{:}-5 a_{1}+8 a_{2}, a_{8}=8 a_{1}+13 a_{2}$, then $5 a_{1}+8 a_{1}=120$, so $a_{1}, a_{2}$ are multiples of 8 and 5 respectively. Let $a_{1}=8 u, a=5 v$, then $u+v-3$, so $u=1, v=2$, or $u=2, v=1$. Since the sequence is increasing, it can only be $u=1, v=2$. Therefore, $a_{1}=8, a_{2}=10$. Thus, $a_{8}=194$.