# How do you find the sum of the infinite geometric series 18-6+2-...?

Jan 11, 2016

$\frac{27}{2}$

#### Explanation:

We must write the general term of this alternating geometric series in the form $a {r}^{n - 1}$, where $a$ is the first term and $r$ is the common ratio between terms.

So in this case $a = 18$ and $r = - \frac{1}{3}$.

Since $| r | = \frac{1}{3} < 1$ it implies the series converges and the sum is given by $\frac{a}{1 - r}$.

$\therefore {\sum}_{n = 1}^{\infty} 18 \cdot {\left(- \frac{1}{3}\right)}^{n - 1} = \frac{18}{1 - \left(- \frac{1}{3}\right)} = \frac{27}{2}$.