Chapter 12 Sequences and Series - 12.4 Find Sums of Infinite Geometric Series - 12.4 Exercises - Mixed Review - Page 825: 52

$a_n=4(2.5)^{n-1}$

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, $a_n=4(2.5)^{n-1}$