## Intermediate Algebra: Connecting Concepts through Application

$a_n=2(5)^{n-1}$
$a_4=250$ and $a_9=781250$ $\frac{a_9}{a_4}=\frac{ar^8}{ar^3}=\frac{781250}{250}=3125=r^5$ $\implies r=(3125)^{\frac{1}{5}}=5.$ $a_4=ar^3=a(5^3)=125a=250$ $\implies a=2$ The $n^{th}$ term is therefore $a_n=2(5)^{n-1}$