Answer
$a_1=6$
$a_2=-2$
$a_3=\frac{2}{3}$
$a_4=-\frac{2}{9}$
$a_5=\frac{2}{27}$
Work Step by Step
The nth term of a geometric sequence:
$a_n=a_1r^{n-1}$
$a_1=6,~~r=-\frac{1}{3}$
$a_2=6(-\frac{1}{3})^{2-1}=6(-\frac{1}{3})=-2$
$a_3=6(-\frac{1}{3})^{3-1}=6(\frac{1}{9})=\frac{6}{9}=\frac{2}{3}$
$a_4=6(-\frac{1}{3})^{4-1}=6(-\frac{1}{27})=-\frac{6}{27}=-\frac{2}{9}$
$a_5=6(-\frac{1}{3})^{5-1}=6(\frac{1}{81})=\frac{6}{81}=\frac{2}{27}$
