Question

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

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

Answer 1

Answer: (D)

First term = a & common ratio = r
Given $S_{\infty }=4\&a_2=\frac{3}{4}$
$\Rightarrow \frac{a}{1-r}=4$...(1)
$\&ar=\frac{3}{4}[\because S_{\infty }=\frac{a}{1-r}\&a_n=a{r}^{n-1}]$
$\Rightarrow a=\frac{3}{4r}$
$\therefore$ Equation (1) becomes $\frac{3}{4r(1-r)}=4$
$\Rightarrow 16r^2-16r+3=0$
$\Rightarrow$ (4r - 3) (4r - 1) = 0
$\Rightarrow r=\frac{3}{4}orr=\frac{1}{4}$
when $r=\frac{1}{4}$ then a= $\frac{3}{4\times \frac{1}{4}}=3$
$\therefore a=3\&r=\frac{1}{4}$