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.
Asked by Vikrant | 30th May, 2012, 11:03: AM
Let x, A1, A2,
An, 2y be in AP.
Then common difference = (2y - x)/ (n + 1)
Ar = x + rd = x + r (2y - x)/ (n + 1) = [x (n + 1) + r(2y - x)]/ (n + 1)
Now 2x, B1, B2,
Bn, y are in AP.
Then common difference = (y - 2x)/ (n + 1)
Br = x + rd = x + r (y - 2x)/ (n + 1) = [x (n + 1) + r(y - 2x)]/ (n + 1)
Since, Ar = Br, 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.
Let x, A1, A2, An, 2y be in AP.
Then common difference = (2y - x)/ (n + 1)
Ar = x + rd = x + r (2y - x)/ (n + 1) = [x (n + 1) + r(2y - x)]/ (n + 1)
Now 2x, B1, B2, Bn, y are in AP.
Then common difference = (y - 2x)/ (n + 1)
Br = x + rd = x + r (y - 2x)/ (n + 1) = [x (n + 1) + r(y - 2x)]/ (n + 1)
Since, Ar = Br, 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.
Answered by | 30th May, 2012, 12:31: PM
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