## Algebra and Trigonometry 10th Edition

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