Given that n AM's are inserted between two sets of numbers a,2 b and.2a,b where a,b∈RSuppose further that mth mean between these sets ofnumbers is same, then the ratio a :b is equal to

A
$(n - m + 1) : m$
B
$(n - m + 1) : n$
C
$n : (n - m + 1)$
D
$m : (n - m + 1)$

Solution:

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

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

According to the given condition,

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

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

Solution:

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