Question

Consider an infinite geometric series with first term $a$ and common ratio $r.$ If the sum is $4$ and the second term is $\frac{3}{4}$, then

• A. $a=2,r=\frac{3}{8}$
• B. $a=\frac{4}{7},r=\frac{3}{7}$
• C. $a=\frac{3}{2},r=\frac{1}{2}$
• D. $a=3,r=\frac{1}{4}$

Answer 1

The correct option is D

$a=3,r=\frac{1}{4}$

Finding the value:

Given sum of infinite geometric series $=4$

$\Rightarrow$ $\frac{a}{1-r}=4$ -------(1)

Second Term$=ar=\frac{3}{4}$

$\Rightarrow$ $a=\frac{3}{4r}$------(2)

Substitute$ain\left(1\right)$, we have

$\frac{3}{4r}\times \frac{1}{1-r}=4$

$\Rightarrow$ $16r^2-16r+3=0$

$\Rightarrow$ $(4r-3)(4r-1)=0$

$\Rightarrow$ $r=\frac{3}{4}orr=\frac{1}{4}$

In option, $r=\frac{1}{4}$is given.

Putting $r=\frac{1}{4}$in equation (2), we have

$a=3$

Hence, Option(D) is correct.