Answer
$a_n=-36(-\frac{1}{2})^n$
$a_1=18$
$a_2=-9$
$a_3=\frac{9}{2}$
$a_4=-\frac{9}{4}$
$a_5=\frac{9}{8}$
Work Step by Step
$a_1=18,~~a_2=-9$
$r=\frac{a_2}{a_1}=\frac{-9}{18}=-\frac{1}{2}$
The nth term of a geometric sequence:
$a_n=a_1r^{n-1}$
$a_n=18(-\frac{1}{2})^{n-1}=18(-\frac{1}{2})^n(-\frac{1}{2})^{-1}=18(-2)(-\frac{1}{2})^n=-36(-\frac{1}{2})^n$
$a_3=-36(-\frac{1}{2})^3=-36(-\frac{1}{8})=\frac{9}{2}$
$a_4=-36(-\frac{1}{2})^4=-36(\frac{1}{16})=-\frac{9}{4}$
$a_5=-36(-\frac{1}{2})^5=-36(-\frac{1}{32})=\frac{9}{8}$
