Given that n AM's are inserted between two sets of numbers a,2 b and.2a,b where $\mathrm{a},\mathrm{b}\in \mathrm{R}$Suppose further that mth mean between these sets of
numbers is same, then the ratio a :b is equal to

$\mathrm{m}\text{ th mean between }\mathrm{a}\text{ and }2\mathrm{b}\text{ is }\mathrm{a}+\frac{\mathrm{m}(2\mathrm{b}-\mathrm{a})}{\mathrm{n}+1}\text{. }$

$\text{ and }\mathrm{m}\text{ th mean between }2\mathrm{a}\text{ and }\mathrm{b}\text{ is }2\mathrm{a}+\frac{\mathrm{m}(\mathrm{b}-2\mathrm{a})}{\mathrm{n}+1}\text{. }$

According to the given condition,

$\begin{array}{l}\mathrm{a}+\frac{\mathrm{m}(2\mathrm{b}-\mathrm{a})}{\mathrm{n}+1}=2\mathrm{a}+\frac{\mathrm{m}(\mathrm{b}-2\mathrm{a})}{\mathrm{n}+1}\\ \Rightarrow \mathrm{m}(2\mathrm{b}-\mathrm{a})=\mathrm{a}(\mathrm{n}+1)+\mathrm{m}(\mathrm{b}-2\mathrm{a})\\ \Rightarrow \mathrm{a}(\mathrm{n}-\mathrm{m}+1)=\mathrm{bm}\\ \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{m}}{\mathrm{n}-\mathrm{m}+1}\end{array}$