5. The increasing sequence of positive integers $a_{1}, a_{2}, a_{3}, \cdots$ satisfies $a_{n+2}=a_{n}+a_{n+1}(n \geqslant 1)$. If $a_{7}=120$, then $a_{8}$ equals
A. 128
B. 168
C. 193
D. 194
5. D Let $a_{1}=a, a_{2}=b$, recursively we can determine the sequence as: $a, b, a+b, a+2b, 2a+3b, 3a+5b, 5a+8b, \cdots$

Therefore,
$$
a_{7}=8b+5a=120 \text{. }
$$

From this, we have $a=\frac{120}{5}-\frac{8b}{5}=24-\frac{8b}{5}$, where $a, b$ are positive integers.
Thus, when $b=5,10,15$, $a=16,8,0$.
$a=0$ does not meet the problem's conditions. Also, $a=16, b=5$ contradicts the assumption of an increasing sequence. Therefore, $a=8, b=10$,
and $a_{8}=a_{7}+a_{6}=8a+13b$, so $a_{8}=8 \cdot 8+13 \cdot 10=194$.
Hence, the answer is D.