Given that $n$ A.M.s are inserted between two sets of numbers $a,2b$ and $2a,b$ where  $a,b\in R$ if the $m^{th}$ mean between these sets of numbers is same, then the ratio $a:b=$

I set

$a\underline{n\text{ A.M's }}2b$

$d=\frac{2b-a}{n+1}$

$m^{th}\text{ mean}$

$\text{Am}=a+md$

$=a+m\left[\frac{2b-a}{n+1}\right]\to \left(1\right)$

II set

2a  $\underset{\_}{n\text{A.M's}}$ b

$d=\frac{b-2a}{n+1}$

$m^{th}$ mean

$\text{Am}=2a+md$

$=2a+m\left[\frac{b-2a}{n+1}\right]\to \left(2\right)$

$\left(1\right)=\left(2\right)$

$a+m\left[\frac{2b-a}{n+1}\right]=2a+m\left[\frac{b-2a}{n+1}\right]$

$\frac{2mb-ma}{n+1}-\frac{mb-2ma}{n+1}=2a-a$

$\begin{array}{l}2mb-ma-mb+2ma=(n+1)a\\ ma+mb=na+a\\ mb=na+a-ma\end{array}$

$mb=a(n+1-m)$

$\frac{a}{b}=\frac{m}{n-m+1}\Rightarrow a:b=m:n-m+1$