Answer
$S_{41}\approx12.5$
Work Step by Step
$\displaystyle \sum_{j=0}^{40}5(\frac{3}{5})^n=5(\frac{3}{5})^0+5(\frac{3}{5})^1+5(\frac{3}{5})^2+...+5(\frac{3}{5})^{40}$
There are 41 terms is this sequence.
$a_1=5(\frac{3}{5})^0=5$
$r=\frac{a_2}{a_1}=\frac{5(\frac{3}{5})^1}{5(\frac{3}{5})^0}=\frac{3}{5}$
$S_n=a_1(\frac{1-r^n}{1-r})$
$S_{41}=5(\frac{1-(\frac{3}{5})^{41}}{1-\frac{3}{5}})=5(\frac{1-(\frac{3}{5})^{41}}{\frac{2}{5}})=\frac{25}{2}[1-(\frac{3}{5})^{41}]\approx12.5$
