Q. If an infinite geometric series the first term is $a$ and common ratio is $r$. If the sum of the series is 4 and the second term is 3/4, then $(a,r)$ is
A) $(\frac{4}{7},\frac{3}{7})$
B) $(2,\frac{3}{8})$
C) $(\frac{3}{2},\frac{1}{2})$
D) $(3,\ \frac{1}{4})$
Explanation
(d)
Since, $\frac{a}{1 - r} = 4\ \ \ \ \Rightarrow a = 4(1 - r)$ …(i)
and $ar = \frac{3}{4}$
$\Rightarrow 4(1 - r)r = \frac{3}{4}$ [from Eq. (i)]
$\Rightarrow 16r^{2} - 16r + 3 = 0\ \ \Rightarrow (4r - 1)(4r - 3) = 0$
$\Rightarrow \ r = \frac{1}{4},\ \frac{3}{4}$
If $r = \frac{1}{4}$, then $a = 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.