Answer
The equation for the $n$th term is
$a_n=486(\frac{1}{3})^{n-1}$
and $a_9=\frac{2}{27}$
Work Step by Step
The given series is
$486,162,54,18,...$
First term $a_1=486$.
Common ratio $r=\frac{162}{486}=\frac{1}{3}$.
Equation for a geometric sequence is
$\Rightarrow a_n=a_1r^{n-1}$
Substitute $486$ for $a_1$ and $\frac{1}{3}$ for $r$.
$\Rightarrow a_n=486(\frac{1}{3})^{n-1}$
$\Rightarrow a_n=486(\frac{1}{3})^{n-1}$
Substitute $9$ for $n$.
$\Rightarrow a_9=486(\frac{1}{3})^{9-1}$
Simplify.
$\Rightarrow a_9=486(\frac{1}{3})^{8}$
$\Rightarrow a_9=\frac{2}{3^3}$.
$\Rightarrow a_9=\frac{2}{27}$.
