Answer
The Sum of the given infinite geometric series = 4
Work Step by Step
The Sum of a infinite geometric series ( if | r | $\lt$ 1 ) is given by
S = $\frac{First term }{1 - common ratio}$ = $\frac{a_{1}}{1 - r}$
The given infinite geometric series
= 3 + $\frac{3}{4}$ + $\frac{3}{4^{2}}$ + $\frac{3}{4^{3}}$ + ......................
= = 3 + $\frac{3}{4}$ + $\frac{3}{16}$ + $\frac{3}{64}$ + ......................
Here
First term $a_{1}$ = 3
common ratio r = $\frac{\frac{3}{64}}{\frac{3}{16}}$ = $\frac{\frac{3}{16}}{\frac{3}{4}}$ = $\frac{\frac{3}{4}}{3}$ = $\frac{1}{4}$
The Sum of the given infinite geometric series = $\frac{a_{1}}{1 - r}$ = $\frac{3}{1 - \frac{1}{4}}$ = $\frac{3}{\frac{3}{4}}$ = 3$\times$$\frac{4}{3}$ = 4
