Answer
$a_{n}=-3(-\frac{2}{3})^{n-1}$
Work Step by Step
{$-3,~2,~-\frac{4}{3},~\frac{8}{9},~-\frac{16}{27},~...$}
$\frac{a_{2}}{a_{1}}=\frac{2}{-3}=-\frac{2}{3}$, $\frac{a_{3}}{a_{2}}=\frac{-\frac{4}{3}}{2}=-\frac{2}{3}$. And so on.
So, $a_{2}=a_{1}(-\frac{2}{3}),~a_{3}=a_{2}(-\frac{2}{3})=a_{1}(-\frac{2}{3})^{2}$
So, $a_{n}=-3(-\frac{2}{3})^{n-1}$
