## Algebra 2 (1st Edition)

$a_n=4(2.5)^{n-1}$
Here, we have $a_n= a_1r^{(n-1)}$ for the Geometric series. We are told that $a_3=25$ and $r=2.5$ Here, we have $a_3=(25) \times (2.5)^{(3-1)}$ or, $a_3=a_1r^2$ Now, $a_1=\dfrac{a_3}{r^2}=\dfrac{25}{(2.5)^2}=4$ Hence, $a_n=4(2.5)^{n-1}$