Find the relation between x and y such that the rth mean between x and 2y may be same as the rth mean between 2x and y, if n means are inserted in each case.

Let x, A₁, A₂, … A_n, 2y be in AP.

Then common difference = (2y - x)/ (n + 1)

A_r = x + rd = x + r (2y - x)/ (n + 1) = [x (n + 1) + r(2y - x)]/ (n + 1)

Now 2x, B₁, B₂, … B_n, y are in AP.

Then common difference = (y - 2x)/ (n + 1)

B_r = x + rd = x + r (y - 2x)/ (n + 1) = [x (n + 1) + r(y - 2x)]/ (n + 1)

Since, A_r = B_r, we have:

[x (n + 1) + r(2y - x)]/ (n + 1) = [x (n + 1) + r(y - 2x)]/ (n + 1)

On simplifying, we get,

ry = x(n - r + 1)

This is the required relation.