# How do you find the sum of the infinite geometric series given 18-12+8-...?

Jun 24, 2018

$\frac{54}{5}$

#### Explanation:

for a GP

$a , a r , a {r}^{2} , \ldots a {r}^{n - 1} , \ldots . .$

the sum to in infinity is

${S}_{\infty} = \frac{a}{1 - r}$

provided
$| r | < 1$

we have

$18 - 12 + 8 - \ldots$

$r = - \frac{12}{18} = - \frac{2}{3}$

$| - \frac{2}{3} | < 1 , \therefore \exists {S}_{\infty}$

${S}_{\infty} = \frac{18}{1 - - \frac{2}{3}}$

=18/((1+2/3)

$= \frac{18}{\frac{5}{3}}$

$= \frac{18 \times 3}{5}$

$= \frac{54}{5}$