Q. If sum of an infinite geometric series is $\frac{4}{3}$ and its 1st term is $\frac{3}{4}$, then its common ration is
A) $\frac{7}{16}$
B) $\frac{9}{16}$
C) $\frac{1}{9}$
D) $\frac{7}{9}$
Explanation
(a)
Given, $S_{\infty} = \frac{4}{3}$ and $a = \frac{3}{4}$
Let $r$ be the common ratio.
$\therefore\frac{a}{1 - r} = \frac{4}{3}$
$\Rightarrow \frac{4}{3} - \frac{4}{3}\ r = \frac{3}{4}$
$\Rightarrow \frac{16 - 9}{12} = \frac{4}{3}r$
$\Rightarrow \frac{7}{12} = \frac{4}{3}r$
$\Rightarrow r = \frac{7}{16}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.