5. The first term of an arithmetic sequence is 9, and the 8th term is 12. How many terms in the first 2015 terms of this sequence are multiples of 3?
【Answer】 288

【Solution】According to the given conditions, it is easy to derive the general term formula of this arithmetic sequence as $a_{n}=a_{1}+(n-1) d=\frac{3 n+60}{7}=\frac{3(n+20)}{7}$. To make it a multiple of 3, it is only necessary to make $\frac{n+20}{7}$ an integer. It is easy to know that when $n=1, 8, 15, \cdots \cdots, 2010$, the condition is satisfied, with a total of $\frac{2010-1}{7}+1=288$ terms meeting the requirement.