Answer
$\dfrac{64}{99}$
Work Step by Step
The sum of a geometric series can be found as:
$S_n=\dfrac{a_1}{1-r}$
where, $a_1$ denotes the first term and the $r$ is common ratio.
We are given that $\overline{0.64}=0.64+0.0064+....=\dfrac{64}{10^2}+\dfrac{64}{10^4}+...$
This can be further written as: $\dfrac{64}{10^2}+\dfrac{64}{10^4}+...=\Sigma_{n=0}^{\infty} \dfrac{64}{10^2} (\dfrac{1}{10^2})^n=\Sigma_{n=0}^{\infty} \dfrac{64}{100} (\dfrac{1}{10^2})^n$
So, we have $a_1=\dfrac{64}{100}$ and $r=\dfrac{1}{100}$
Thus, the total sum of two geometric series is:
$S_n=\dfrac{64/100}{1-\dfrac{1}{100}}=\dfrac{64}{99}$
