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 $mth$ mean between these sets of numbers are same, then the ratio $a:b$ equals

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

Answer 1

The correct option is C $m:n-m+1$

$a...n$ A.M.s$........2b$
$d=\frac{2b-a}{n+1}$
$A_m=a+m\left(\frac{2b-a}{n+1}\right)...\left(1\right)$
$2a.........n$ A.M's $...........b$
$d=\frac{b-2a}{n+1}$
$A_m=2a+m\left(\frac{b-2a}{n+1}\right)...\left(2\right)$
equating (1) and (2)
$a=\frac{m}{n+1}(b+a)\Rightarrow \frac{a}{b}=\frac{m}{n-m+1}$