Answer
$a_n=-3(-\frac{1}{3})^{n-1}.$
Work Step by Step
We know that $a_n=a_1r^{n-1}.$
where $r$=common ration (we can get this by dividing any two consecutive terms) and $a_1$= the first term
Here $a_1=-3$ and $r=\frac{a_2}{a_1}=\dfrac{1}{-3}=-\frac{1}{3}$
Hence, here we have
$a_n=-3(-\frac{1}{3})^{n-1}.$