## Precalculus (6th Edition) Blitzer

$1536$
we have $a_1=3$ and $r=2$ For a geometric sequence $a_n=a_1r^{n-1}$ Thus, $a_n=3(2)^{n-1}$ Now, $a_{10}=3(2)^{10-1}=3(2)^9=1536$