Chapter 9 - Conic Sections, Sequences, and Series - 9.4 Geometric Sequences - 9.4 Exercises - Page 737: 32

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

$a_6=288$ 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 $a_n=9(2)^{n-1}$