#### Answer

$a_n=9(2)^{n-1}$

#### Work Step by Step

$a_6=288$ and $a_{13}=36864$
$\frac{a_{13}}{a_6}=\frac{ar^12}{ar^5}=\frac{36864}{288}=128=r^7$
$\implies r=(128)^{\frac{1}{7}}=2.$
$a_6=ar^5=a(2^5)=32a=288$
$\implies a=9$
The $n^{th}$ term is therefore
$a_n=9(2)^{n-1}$