## Intermediate Algebra: Connecting Concepts through Application

$a_n=(-4)\left(\frac{1}{2}\right)^{n-1}$
$-4,-2,-1,-\frac{1}{2},-\frac{1}{4},-\frac{1}{8},...$ $\frac{a_2}{a_1}=\frac{-2}{-4}=\frac{1}{2}=r$ $ra_2=\frac{1}{2}\times(-2)=-1=a_3$ $ra_3=\frac{1}{2}\times(-1)=-\frac{1}{2}=a_4$ and so on. The sequence is geometric with ratio $r=\frac{1}{2}$. The first term is $a=-4$ and so the $n^{th}$ term is given by $a_n=(-4)\left(\frac{1}{2}\right)^{n-1}$