Answer
$\displaystyle \sum_{j=0}^{50}10(\frac{2}{3})^{n-1}\approx45$
Work Step by Step
$\displaystyle \sum_{j=0}^{50}10(\frac{2}{3})^{n-1}=10(\frac{2}{3})^{-1}+10(\frac{2}{3})^{0}+10(\frac{2}{3})^1+...+10(\frac{2}{3})^{49}$
There are 51 terms is this sequence.
$a_1=10(\frac{2}{3})^{-1}=10(\frac{3}{2})=15$
$r=\frac{a_2}{a_1}=\frac{10(\frac{2}{3})^0}{10(\frac{2}{3})^{-1}}=\frac{2}{3}$
$S_n=a_1(\frac{1-r^n}{1-r})$
$S_{51}=15(\frac{1-(\frac{2}{3})^{51}}{1-(\frac{2}{3})})=15(\frac{1-(\frac{2}{3})^{51}}{\frac{1}{3}})\approx45$
