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(D) Let $d_1$ be the common difference for the first set $(a, 2b)$. The $(n+2)^{th}$ term is $2b = a + (n+1)d_1$,so $d_1 = \frac{2b-a}{n+1}$.The $m^{t

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$m : n-m+1$

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(D) Let $d_1$ be the common difference for the first set $(a, 2b)$. The $(n+2)^{th}$ term is $2b = a + (n+1)d_1$,so $d_1 = \frac{2b-a}{n+1}$.The $m^{th}$ arithmetic mean is $A_m = a + m d_1 = a + m \left( \frac{2b-a}{n+1} \right)$.Let $d_2$ be the common difference for the second set $(2a, b)$. The $(n+2)^{th}$ term is $b = 2a + (n+1)d_2$,so $d_2 = \frac{b-2a}{n+1}$.The $m^{th}$ arithmetic mean is $A'_m = 2a + m d_2 = 2a + m \left( \frac{b-2a}{n+1} \right)$.Equating the two means: $a + m \left( \frac{2b-a}{n+1} \right) = 2a + m \left( \frac{b-2a}{n+1} \right)$.Multiplying by $(n+1)$: $a(n+1) + m(2b-a) = 2a(n+1) + m(b-2a)$.$an + a + 2bm - am = 2an + 2a + bm - 2am$.Rearranging terms: $2bm - bm - am + 2am = 2an - an + 2a - a$.$bm + am = an + a$.$m(b+a) = a(n+1)$.Thus,$\frac{a}{b} = \frac{m}{n+1-m}$.

Given that $n$ arithmetic means are inserted between two sets of numbers $(a, 2b)$ and $(2a, b)$,where $a, b \in \mathbb{R}$. Suppose the $m^{th}$ means between these sets are equal,then the ratio $a : b$ is equal to:

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