Question

Given that n numbers of A.Ms are inserted between two sets of numbers $a,2b$ and $2a,b$ where $a,b\in R.$
Suppose further that the $m^{th}$ means between these sets of numbers are same, then the ratio $a:b$ equals

• A. $n–m+1:m$
• B. $n–m+1:n$
• C. $n:n–m+1$
• D. $m:n–m+1$

Answer 1

The correct option is D $m:n–m+1$

$a\cdots nA.M^{\prime }s\cdots 2b$
So total number of terms in the series is $n+2$
So common difference
$d=\frac{2b-a}{n+1}$
$A_m=a+m\left(\frac{2b-a}{n+1}\right)\cdots \left(1\right)$
simillarly
$2a\cdots nA.M^{\prime }s\cdots bd=\frac{b-2a}{n+1}$
$A_m=2a+m\left(\frac{b-2a}{n+1}\right)\cdots \left(2\right)$
equating $\left(1\right)\&\left(2\right)$
$a=\frac{m}{n+1}(b+a)\Rightarrow \frac{a}{b}=\frac{m}{n-m+1}$