Answer
$a_n=-4(-\frac{3}{2})^n$
Work Step by Step
$a_1=5$
$r=\frac{a_2}{a_1}=\frac{a_3}{a_2}$
$r=\frac{-9}{6}=\frac{\frac{27}{2}}{-9}=-\frac{3}{2}$
$a_n=a_1r^{n-1}$
$a_n=6(-\frac{3}{2})^{n-1}=6(-\frac{3}{2})^n(-\frac{3}{2})^{-1}=6(-\frac{3}{2})^n(-\frac{2}{3})=-4(-\frac{3}{2})^n$
