Question

If sum of an infinite geometric series is $\frac{4}{3}$ and its Ist term is $\frac{3}{4}$, then its common ratio is

• A. $\frac{7}{16}$
• B. $\frac{9}{16}$
• C. $\frac{1}{9}$
• D. $\frac{7}{9}$

Answer 1

The correct option is A $\frac{7}{16}$

We know that sum of infinite geometric series $=\frac{a}{1-r}$
Where $a=\text{first term}$ and $r=\text{common ratio}$.
$\frac{a}{1-r}=\frac{4}{3}$
Then, $\frac{\frac{3}{4}}{1-r}=\frac{4}{3}\Rightarrow r=1-\frac{9}{16}=\frac{7}{16}$