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Find the relation between x and y in order that the $r^{th}$ mean between x and 2y may be the same as the $r^{th}$ mean between $2 x$ and $y, n$ means being inserted in each case.

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y = (\frac{n - r + 1}{r}) x

y = (\frac{n - r + 2}{r}) x

y = (\frac{n - r + 3}{r}) x

y = (\frac{n - r + 4}{r}) x

answer is A.

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Detailed Solution

In case of n means inserted between two terms a and b, then the common difference of will be,

d = \frac{b - a}{n + 1}

Here, the value of

r^{th} mean,

= a + rd

Here, we have two terms x and y. So,

x + r \times (\frac{2 y - x}{n + 1}) = 2 x + r \times (\frac{y - 2 x}{n + 1})

\Rightarrow (n + 1) x + 2 ry - rx = 2 (n + 1) x + ry - 2 rx

\Rightarrow y = (\frac{n - r + 1}{r}) x

This is the required relation.
Thus, the correct option is 1.

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