Some sort of Sum

It is known that

$\sum_{x = 0}^{\infty}{ar^x} = \frac{a}{1 - r}$

We already have the common ratio as $r$ , all we need now is the first term $a$ . We only need to make $x$ equal to $0$ to find $a$ .

$r^{0 - 1} = r^{-1} = \frac{1}{r}$

We'll then insert the values into the equation

$\frac{4}{r} = \frac{\frac{1}{r}}{1 - r}$

Solving for $r$

$\frac{4}{r} = \frac{1}{r - r^2}$

$\frac{4r - 4r^2}{r} = 1$

$4 - 4r = 1$

$-4r = -3$

$4r = 3$

$r = \frac{3}{4}$