given that n arithmetic means are inserted between two sets of numbers a, 2b and 2a, b where a, b are real numbers . suppose further that mth mean between these two sets of numbers is same then the ratio a : b equals

1] n-m+1 : m 2] n-m+1 : n 3] m : n-m+1 4] n : n-m+1

since n airthmetic means are inserted between a , 2b.
total number of terms = n+2
let the common difference be  d.
$$\begin{gathered} 2b=a+(n+1)d \\ d=\frac{2b-a}{n+1}.......\left(1\right) \end{gathered}$$
1st airthmetic mean = a+d = $a+\frac{2b-a}{n+1}$
mth airthmetic mean = a+md = $a+m*\frac{2b-a}{n+1}$
similarly m th mean for the other pair 2a and b is
$2a+m*\frac{b-2a}{n+1}$
since both the mth mean are equal:
$$\begin{gathered} a+m*\frac{2b-a}{n+1}=2a+m*\frac{b-2a}{n+1} \\ \frac{m}{n+1}*(2b-a-b+2a)=a \\ \frac{m}{n+1}*(b+a)=a \\ m(b+a)=a(n+1) \\ mb+ma=a(n+1) \\ a(n-m+1)=mb \\ \frac{a}{b}=\frac{m}{n-m+1} \end{gathered}$$

hope this helps you