To demonstrate that $7 \mid a_{n}=\frac{3^{n+1}-1}{2}$ is equivalent to showing that $3^{n+1}-1 \equiv 0 \bmod 7$. However $3^{n+1}-1=0 \bmod 7 \Longrightarrow n+1 \equiv 0 \bmod 6$, hence the values of $n \leq 2012$ are $5,11,17, \ldots, 2009$, which is in total $$\frac{2009+1}{6}=335$$ possible terms.