10.4 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.
(E) 210.
(43rd American High School Mathematics Examination, 1992)
[Sol] Let $a_{1}=a, a_{2}=b$, recursively we know the sequence is: $a, \quad b, \quad a+b, \quad a+2 b, 2 a+3 b, 3 a+5 b, \quad 5 a+8 b$, thus
$$
a_{7}=8 b+5 a=120 \text {. }
$$

From this, we have $a=\frac{120}{5}-\frac{8 b}{5}=24-\frac{8 b}{5}$, where $a, b$ are positive integers.
Therefore, 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.
Thus, $a=8, b=10$,
and $a_{8}=a_{7}+a_{6}=8 a+13 b$, so $a_{8}=8 \cdot 8+13 \cdot 10=194$. Therefore, the answer is $(D)$.