Precalculus (6th Edition) Blitzer

$a_{n}=16(\dfrac{1}{4})^{n-1}$ and $a_{12}=\dfrac{1}{(4)^9} =\dfrac{1}{262144}$
Here, $a_1=16$ and $d=\dfrac{1}{4}$ $a_{n}=16(\dfrac{1}{4})^{n-1}$ Now, plug in: $n=12$ $a_{12}=16(\dfrac{1}{4})^{12-1}$ and $a_{12}=\dfrac{1}{(4)^9} =\dfrac{1}{262144}$ So, $a_{n}=16(\dfrac{1}{4})^{n-1}$ and $a_{12}=\dfrac{1}{(4)^9} =\dfrac{1}{262144}$