If $n$  arithmetic means are inserted between two sets of numbers $a,2b$  and $2a,b$  where $a,b\in R$ . Suppose that $m^{\text{th}}$  arithmetic mean between these two sets of numbers is same, then the ratio $a:b$  equals

$m^{th}A.M$  means ${\left(m+1\right)}^{th}$  term

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

$\Rightarrow \left(n+1\right)a+m\left(2b-a\right)=2a\left(n+1\right)+m\left(b-2a\right)$

$\Rightarrow na+a+2bm-ma=2an+2a+mb-2am$

$\Rightarrow na+a-ma-2an-2a+2am=mb-2bm$

$\Rightarrow a\left[n+1-m-2n-2+2m\right]=b\left(m-2m\right)$

$\Rightarrow a\left(m-n-1\right)=b\left(-m\right)\Rightarrow \frac{a}{b}=\frac{-m}{m-n-1}=\frac{m}{1+n-m}$