using a base 3 decimal to express as a base 10 fraction using geometric series 
Express $0.\overline{21}_3$ as a base 10 fraction in reduced form.

So I was able to solve it by setting $x=\overline{.21}$, but the solution also briefly mentioned another way using the geometric series:

A quick way to get the answer by using the geometric series is: $(0.212121 \ldots)_3 = \frac{7}{9} + \frac{7}{81} + \frac{7}{729} + \dots = \frac{7}{8}.$

However, I'm having a hard time understanding how to actually use the geometric series (the above answer is not clear to me). 

You have
$(0.212121 \ldots)_3 = \frac{7}{9} + \frac{7}{81} + \frac{7}{729} + \dots$
and for $0<a<1$
$1+a+a^2+\ldots = \dfrac{1}{1-a}$
from which 
$\begin {align}
(0.212121 \ldots)_3 &= 7\left(\frac{1}{9} + \frac{1}{81} + \frac{1}{729} + \dots 
 \right)\\
 &= 7\left(\frac{1}{1-\frac{1}{9}} -1 \right)\\
 &= 7\left(\frac{9}{8} -1 \right)\\
 &= \frac{7}{8} \\
\end{align}$

You can also (perhaps this is what you actually did) multiply by $100_3 = 9_{\text{dec}}$ to see that for $x=0.\overline{21}_3$, $9x-x = 21_3 = 7_{\text{dec}}$ and solve from there.
