Question

If sum of an infinite geometric series is $\frac{4}{5}$ and its first term is $\frac{3}{4}$, then its common ratio is

• A. $\frac{7}{16}$
• B. $\frac{9}{16}$
• C. $\frac{1}{9}$
• D. none of these

Answer 1

The correct option is D

none of these

Explanation for the correct option:

Given the sum of infinite geometric series is $\frac{4}{5}$ and the first term is $\frac{3}{4}$.

The general form of the infinite geometric series is $a+a_{}r+a_{}{r}^2+a_{}{r}^3+\dots ,$where $a$ is the first term and $r$ is the common ratio.

Thus, the formula for the sum of infinite $\left(S_{\infty }\right)$:

$\therefore S_{\infty }=\frac{a}{1-r}$

Now,

$\begin{array}{c}\Rightarrow \frac{a}{1-r}=\frac{4}{5}\left[\because S_{\infty }=\frac{4}{5}\right]\\ \Rightarrow \frac{3}{4\left(1-r\right)}=\frac{4}{5}\left[\because a=\frac{3}{4}\right]\\ \Rightarrow 15=16-16r\\ \Rightarrow 16r=1\\ \Rightarrow r=\frac{1}{16}\end{array}$

Hence option (D) is the answer.