## Intermediate Algebra: Connecting Concepts through Application

$a_n=625\left(\frac{1}{5}\right)^{n-1}$
$625, 125, 25, 5, 1, 0.2, . . .$ $\frac{a_2}{a_1}=\frac{125}{625}=\frac{1}{5}=r$ $ra_2=\frac{1}{5}\times125=25=a_3$ $ra_3=\frac{1}{5}\times25=5=a_4$ and so on. The sequence is geometric with ratio $r=\frac{1}{5}$. The $n^{th}$ term is given by $a_n=625\left(\frac{1}{5}\right)^{n-1}$