## Algebra 2 (1st Edition)

$a_n=66 (\dfrac{1}{2})^{n-1}$
From the previous part it can be seen that $a_1=66; r=\dfrac{1}{2}$. Thus, the series is geometric. So, we have $a_n=a_1r^{n-1}$ But $a_1=66; r=\dfrac{1}{2}$ Then $a_n=66 (\dfrac{1}{2})^{n-1}$