Answer
$$a_n = (-1)^n\frac{2^{n-1}}{3^{n-2}}.$$
Work Step by Step
Our first term is $a_1 = -3$. Because the sequence alternates, we have a $(-1)^{n}$ in the sequence. Each term also changes by a factor of $\frac{2}{3}$, so we introduce a factor of $(\frac{2}{3})^n$. Putting together the pieces with the fact that $a_1 = -3$, we have:
$$a_n = (-1)^n3(\frac{2}{3})^{n-1} = (-1)^n\frac{2^{n-1}}{3^{n-2}}$$
