Answer
$S_{13}\approx-3745.102$
Work Step by Step
$\dfrac{a_{n+1}}{a_n}=\dfrac{-2500\left(\frac{3}{5}\right)^{n+1}}{-2500\left(\frac{3}{5}\right)^n}=\dfrac{3}{5}$
Since the ratio between consecutive terms is constant, the given sequence is geometric.
The first term is $a_{1}=-2500\left(\frac{3}{5}\right)^{1}=-1500$.
The common ratio is $r=\frac{3}{5}$.
The formula for the sum of the first $n$ terms is:
$S_{n}=\frac{a_{1}(1-r^{n})}{1-r}$
$S_{13}=\frac{-1500\left[1-\left(\frac{3}{5}\right)^{13}\right]}{1-\frac{3}{5}}\approx-3745.102$
