Answer
$S_n=12(\frac{1-(-\frac{1}{12})^n}{1-(-\frac{1}{12})})$.
Work Step by Step
$\frac{a_3}{a_2}=\frac{\frac{1}{12}}{-1}=\frac{a_2}{a_1}=\frac{-1}{12}=-\frac{1}{12}=r$, thus this is a geometric sequence with a common ratio of $-\frac{1}{12}$
The nth term of a geometric sequence can be obtained by the following formula: $a_n=a_1(r^{n-1})$, where $a_1$ is the first term and $r$ is the common ratio.
The sum of a geometric sequence until $n$ can be obtained by the formula $S_n=a_1(\frac{1-r^n}{1-r})$ where $a_1$ is the first term and $r$ is the common ratio. Hence here the sum is: $S_n=12(\frac{1-(-\frac{1}{12})^n}{1-(-\frac{1}{12})})$.
