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)
$\& \, ar \, = \frac{3}{4} \, \, [\because \, S_{\infty} \, =\frac{a}{1-r} \, \& \, \, a_n \, = ar^{n-1}]$
$\Rightarrow \, \, a\, = \frac{3}{4r}$
$\therefore $ Equation (1) becomes $\frac{3}{4r(1-r)}=4$
$\Rightarrow \, \, 16r^2-16r+3=0$
$\Rightarrow $ (4r - 3) (4r - 1) = 0
$\Rightarrow \, \, r = \frac{3}{4} \, \, or \, \, r=\frac{1}{4}$
when $r=\frac{1}{4}$ then a= $\frac{3}{4\times \frac{1}{4}}=3$
$\therefore \, \, a=3 \, \& \, r=\frac{1}{4}$