## Intermediate Algebra: Connecting Concepts through Application

$a_n=2(-4)^{n-1}$
$2, -8, 32, -128, 512, -2048, . . .$ $\frac{a_2}{a_1}=\frac{-8}{2}=-4=r$ $ra_2=-4\times-8=32=a_3$ $ra_3=-4\times32=-128=a_4$ and so on. The sequence is geometric with common ratio $r=-4$ The $n^{th}$ term of the sequence is given by $a_n=2r^{n-1}=2(-4)^{n-1}$