we can always get the common ratio of a geometric sequence by simply dividing the current term by the previous term, in this case for the sake of simplicity let's divide 2/3 by 2, so
[tex]\cfrac{2}{3}\div 2\implies \cfrac{2}{3}\div \cfrac{2}{1}\implies \cfrac{2}{3}\cdot \cfrac{1}{2}\implies \cfrac{1}{3}\impliedby \textit{common ratio}[/tex]
so the sequence more or less looks like [tex]2~~,~~\cfrac{2}{3}~~,~~\cfrac{2}{9}~~,~~\cfrac{2}{81}~~,~~\cfrac{2}{243}[/tex]
so we'd want the sum of the first 5 terms
[tex]\qquad \qquad \textit{sum of a finite geometric sequence} \\\\ \displaystyle S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ r=\textit{common ratio} \end{cases} \\\\\\ S_5=2\left( \cfrac{1~~ - ~~(\frac{1}{3})^5}{1~~ - ~~\frac{1}{3}} \right)\implies S_5=2\left( \cfrac{~~\frac{242}{243}~~}{\frac{2}{3}} \right)\implies S_5=\cfrac{242}{81}\implies S_5=2\frac{80}{81}[/tex]
