# How do you find the sum of the infinite geometric series 12(-1/4)^(n - 1)?

The sum of the infinite series $a + a r + a {r}^{2} + a {r}^{3} + \ldots + a {r}^{n - 1} + \ldots$ ia $\frac{a}{1 - r}$, for $| r | < 1$.
Here, a = 12 and $r = - \frac{1}{4}$
Answer = $12 \left(1 + \frac{1}{4}\right) = \frac{48}{5}$..