Answer
$a_n=66 (\dfrac{1}{2})^{n-1}$
Work Step by Step
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
$a_n=66 (\dfrac{1}{2})^{n-1}$
