5. The first term of an arithmetic sequence is 9, and the 8th term is 12. In the first 2015 terms of this sequence, $\qquad$ 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.