Answer
$422$
Work Step by Step
$\sum ^{5}_{j=1}243\left( \dfrac {2}{3}\right) ^{j}=\dfrac {a_{1}\left( r^{n}-1\right) }{r-1};a_{1}=243\times \left( \dfrac {2}{3}\right) ^{1}=162;r=\dfrac {2}{3}\Rightarrow S_{n}=\dfrac {162\times \left( \left( \dfrac {2}{3}\right) ^{5}-1\right) }{\dfrac {2}{3}-1}=\dfrac {162\times \left( \dfrac {-211}{243}\right) }{-\dfrac {1}{3}}=422$
