Chapter 9 - Conic Sections, Sequences, and Series - 9.4 Geometric Sequences - 9.4 Exercises - Page 738: 36

$a_n=8\left(-3\right)^{n-1}$

$a_6=-1944$ and $a_{9}=52488$ $\frac{a_{9}}{a_6}=\frac{ar^{8}}{ar^5}=\frac{52488}{-1944}=-27=r^3$ $\implies r=\left(-27\right)^{\frac{1}{3}}=-3.$ $a_6=ar^5=a\left(-3\right)^5=a\left(-243\right)=-1944=8$ $\implies a=8$ The $n^{th}$ term is therefore $a_n=8\left(-3\right)^{n-1}$