Given that n A.M.'s are inserted between | vedclass

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(d) ${m^{th}}$ mean between $a,\;2b$ is $a + \frac{{m(2b - a)}}{{n + 1}}$ &hellip;&hellip;$(i)$

and ${m^{th}}$ mean between $2a,\;b$ is $2a + \frac

Vedclass · Accepted Answer

$m:n - m + 1$

Vedclass · Answer

(d) ${m^{th}}$ mean between $a,\;2b$ is $a + \frac{{m(2b - a)}}{{n + 1}}$ &hellip;&hellip;$(i)$

and ${m^{th}}$ mean between $2a,\;b$ is $2a + \frac{{m(b - 2a)}}{{n + 1}}$......$(ii)$

Accordingly, $a + \frac{{m(2b - a)}}{{n + 1}} = 2a + \frac{{m(b - 2a)}}{{n + 1}}$

$ \Rightarrow $ $m(2b - a) = a(n + 1) + m(b - 2a)$

$ \Rightarrow $ $a(n - m + 1) = bm$

$ \Rightarrow $ $\frac{a}{b} = \frac{m}{{n - m + 1}}$.

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 ${m^{th}}$ mean between these sets of numbers is same, then the ratio $a:b$ equals

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