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(D) The $m^{th}$ arithmetic mean $(A.M.)$ between two numbers $x$ and $y$ when $n$ means are inserted is given by $A_m = x + \frac{m(y - x)}{n + 1}$.F

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$m : n - m + 1$

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(D) The $m^{th}$ arithmetic mean $(A.M.)$ between two numbers $x$ and $y$ when $n$ means are inserted is given by $A_m = x + \frac{m(y - x)}{n + 1}$.For the first set $a, 2b$,the $m^{th}$ mean is $A_m = a + \frac{m(2b - a)}{n + 1}$.For the second set $2a, b$,the $m^{th}$ mean is $A'_m = 2a + \frac{m(b - 2a)}{n + 1}$.Given that $A_m = A'_m$,we have:$a + \frac{m(2b - a)}{n + 1} = 2a + \frac{m(b - 2a)}{n + 1}$Subtract $a$ from both sides:$\frac{m(2b - a)}{n + 1} = a + \frac{m(b - 2a)}{n + 1}$Multiply by $(n + 1)$:$m(2b - a) = a(n + 1) + m(b - 2a)$$2bm - am = an + a + bm - 2am$Rearrange terms to group $a$ and $b$:$2bm - bm - am + 2am = a(n + 1)$$bm + am = a(n + 1)$$b(m) = a(n + 1 - m)$Therefore,the ratio $\frac{a}{b} = \frac{m}{n - m + 1}$.

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

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