Question

If sum of an infinite geometric series is $\frac{4}{3}$ and $1^{\mathrm{st}}$ 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

$\frac{7}{16}$

Explanation for the correct option:

Sum of infinite GP:

$S_{\infty }=\frac{a}{1-r},\left|r\right|<1$

Where $a$ is first term

$\begin{array}{l}\mathrm{First}\mathrm{term}a=\frac{3}{4}\\ \frac{\frac{3}{4}}{\left(1-r\right)}=\frac{4}{3}\\ \Rightarrow 1-r=\frac{9}{16}\\ \Rightarrow r=1-\frac{9}{16}\\ \Rightarrow r=\frac{7}{16}\end{array}$ $\left[\because S_{\infty }=\frac{a}{1-r},\left|r\right|<1\right]$

Hence, option (A) is correct