Q. Consider an infinite geometric series with 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) $a = 3,\ r = \frac{1}{4}$
Explanation
(d)
Given that, sum of geometric series = 4 second term $= \frac{3}{4}$
$\Rightarrow \ \frac{a}{1 - r} = 4,\ ar = \frac{3}{4}\ \Rightarrow r = \frac{3}{4a}$
$\therefore\ \frac{a}{1 - \frac{3}{4a}} = 4\ \Rightarrow \ \frac{4a^{2}}{4a - 3} = 4$
$\Rightarrow \ (a - 1)(a - 3) = 0\ \Rightarrow a = 1\ or\ 3$
When $a = 1,\ r = \frac{3}{4}$
When $a = 3,\ r = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.