Answer
$a_n=390625\left(-\frac{2}{5}\right)^{n-1}$
Work Step by Step
$a_7=1600$ and $a_{12}=-16.384$
$\frac{a_{12}}{a_7}=\frac{ar^{11}}{ar^{6}}=r^5=\frac{-16.384}{1600}=-\frac{32}{3125}$
$\implies r=\left(-\frac{32}{3125}\right)^{\frac{1}{5}}=-\frac{2}{5}$
$a_7=ar^6=a\left(-\frac{2}{5}\right)^{6}=a\frac{64}{15625}=1600$
$\implies a=390625$
The $n^{th}$ term is therefore given by
$a_n=390625\left(-\frac{2}{5}\right)^{n-1}$
