Answer
for $r=\frac{2}{3}$:
$a_1=9$
$a_2=6$
$a_3=4$
$a_4=\frac{8}{3}$
$a_5=\frac{16}{9}$
for $r=\frac{2}{3}$:
$a_1=9$
$a_2=-6$
$a_3=4$
$a_4=-\frac{8}{3}$
$a_5=\frac{16}{9}$
Work Step by Step
$a_1=9,~~a_3=4$
The nth term of a geometric sequence:
$a_n=a_1r^{n-1}$
$a_3=a_1r^{3-1}$
$4=9r^2$
$\frac{4}{9}=r^2$
$r=\frac{2}{3}$ or $r=-\frac{2}{3}$
for $r=\frac{2}{3}$:
$a_2=a_1+r^{2-1}=9(\frac{2}{3})=6$
$a_3=4$
$a_4=a_1+r^{4-1}=9(\frac{2}{3})^3=9(\frac{8}{27})=\frac{8}{3}$
$a_5=a_1+r^{5-1}=9(\frac{2}{3})^4=9(\frac{16}{81})=\frac{16}{9}$
for $r=\frac{2}{3}$:
$a_2=a_1+r^{2-1}=9(-\frac{2}{3})=-6$
$a_3=4$
$a_4=a_1+r^{4-1}=9(-\frac{2}{3})^3=9(-\frac{8}{27})=-\frac{8}{3}$
$a_5=a_1+r^{5-1}=9(-\frac{2}{3})^4=9(\frac{16}{81})=\frac{16}{9}$
