# The first term of an infinite geometric series is -8, and its sum is -13 1/3, how do you find the first four terms of the series?

##### 1 Answer
Feb 25, 2017

$- 8 , - \frac{16}{5} , - \frac{32}{25} , - \frac{64}{125}$

#### Explanation:

We need to find the common ratio first.

$\text{ for the GP } a , a r , a {r}^{2} , \ldots . a {r}^{n - 1}$

$a = - 8 , {S}_{\infty} = - 13 \frac{1}{3} = - \frac{40}{3}$

the formula for sum to infinity:$\text{ } {S}_{\infty} = \frac{a}{1 - r} , | r | < 1$

in this case$\text{ } - \frac{40}{3} = - \frac{8}{1 - r}$

${\cancel{- 40}}^{5} \left(1 - r\right) = 3 \times \cancel{- 8}$

$5 - 5 r = 3$

$2 = 5 r \implies r = \frac{2}{5}$

$a = - 8$

$a r = - 8 \times \frac{2}{5} = - \frac{16}{5}$

$a {r}^{2} = - \frac{16}{5} \times \frac{2}{5} = - \frac{32}{25}$

$a {r}^{3} = - \frac{32}{25} \times \frac{2}{5} = - \frac{64}{125}$

first four terms

$- 8 , - \frac{16}{5} , - \frac{32}{25} , - \frac{64}{125}$