Q. Given that $n$ AM’s are inserted between two sets of numbers $a,\ 2b\ and\ 2a,\ b\ where\ a,\ b \in R.$ Suppose further that $m$th mean between these sets of numbers is same, then the ratio $a\ :b$ equals
A) $(n - m + 1):m$
B) $(n - m + 1):n$
C) $n\ :(n - m + 1)$
D) $m\ :(n - m + 1)$
Explanation
(d)
$mth$ mean between $a,2b$ is $a + \frac{m(2b - a)}{n + 1}$ and $m$th mean between $2a,\ b\ is\ 2a + \frac{m(b - 2a)}{n + 1}$.
According to given condition
$a + \frac{m(2b - a)}{n + 1} = 2a + \frac{m(b - 2a)}{n + 1}$
$\Rightarrow \ \ m(2b - a) = a(n + 1) + m(b - 2a)$
$\Rightarrow \ \ a(n - m + 1) = bm$
$\Rightarrow \ \frac{a}{b} = \frac{m}{n - m + 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.