Answer
$S_{12}\approx449.999$
Work Step by Step
$\dfrac{a_{n+1}}{a_n}=\dfrac{900\left(\frac{1}{3}\right)^{n+1}}{900\left(\frac{1}{3}\right)^n}=\dfrac{1}{3}$
Since the ratio between consecutive terms is constant, the given sequence is geometric.
The first term is $a_{1}=900(\frac{1}{3})^{1}=300$.
The common ratio is $r=\frac{1}{3}$.
The formula for the sum of the first $n$ terms is:
$S_{n}=\frac{a_{1}(1-r^{n})}{1-r}$
$S_{12}=\frac{300\left[1-\left(\frac{1}{3}\right)^{12}\right]}{1-\frac{1}{3}}\approx449.999$
