## Precalculus (10th Edition)

$a_n=4(\frac{1}{4})^{n-1}.$
We know that $a_n=a_1r^{n-1}.$ where $r$=common ration (we can get this by dividing any two consecutive terms) and $a_1$= the first term Here $a_1=4$ and $r=\frac{a_2}{a_1}=\frac{1}{4}$ Hence, here we have $a_n=4(\frac{1}{4})^{n-1}.$