# How do you find the sum of the infinite geometric series 15 - 3 + 0.6 - 0.12 + . . .?

Feb 3, 2016

${S}_{\infty} = 12.5$

#### Explanation:

Given is the infinite geometric series $15 - 3 + 0.6 - 0.12 + . . .$
The first term is ${T}_{1} = a = 15$, second term ${T}_{2} = - 3$

Since it's a geometric series, we have to divide the second term by the first term, so ${T}_{2} / {T}_{1} = - {\cancel{3}}^{1} / {\cancel{15}}^{5} = - \frac{1}{5}$
So, ${T}_{2} / {T}_{1} = r = \frac{- 1}{5}$

The formula for the infinite geometric series is
${S}_{\infty} = \frac{a}{1 - r}$ for $| r | < 1$
We have all the required terms, so by replacing them for their respective values we have
${S}_{\infty} = \frac{15}{1 - \left(- \frac{1}{5}\right)} = \frac{15}{1 + \frac{1}{5}} = \frac{15}{\frac{6}{5}} = {\cancel{15}}^{5} \cdot \frac{5}{\cancel{6}} ^ 2 = 5 \cdot \frac{5}{2} = \frac{25}{2}$

So now you know how I got that answer up there.