The increasing sequence of positive integers $a_1,a_2,a_3,\cdots$ has the property that
\[a_{n+2}=a_n+a_{n+1} \text{  for all } n\ge 1.\]
If $a_7=120$, then $a_8$ is
$\text{(A) } 128\quad \text{(B) } 168\quad \text{(C) } 193\quad \text{(D) } 194\quad \text{(E) } 210$
$\fbox{D}$ Let $a_{1} = a, a_{2} = b$, so $5a + 8b = 120$. Now $8b$ and $120$ are divisible by $8$, so $5a$ is divisible by 8, so $a$ is divisible by 8. It's now easy to try the multiples of $8$ to get that $a = 8, b=10$ (all the other possibilities violate the condition $a < b$, which comes from the fact that the sequence is increasing). Hence $a_8 = 8a + 13b = 8 \times 8 + 13 \times 10 = 194$.