Show that x2 + xy + y2, z2 + zx + x2 and y2 + yz + z2 are in consecutive terms of an A.P., if x, y and z are in A.P.

Since, x2 + xy + y2, z2 + zx + x2 and y2 + yz + z2 are in AP


(z2 + zx + x2) - (x2 + xy + y2) = (y2 + yz + z2) - (z2 + zx + x2)


Let d = common difference,


Sicex, y, z are in AP


Y = x + d and x = x + 2d


Therefore the above equation becomes,


= (x + 2d)2 + (x + 2d)x - x(x + d) - (x + d)2 = (x + d)2 + (x + d)(x + 2d) - (x + 2d)x – x2


= x2 + 4xd + 4d2 + x2 + 2xd – x2 – xd – x2 - 2xd – d2 = x2 + 2dx + d2 + x2 + 2dx + xd + 2d2 – x2 - 2dx – x2



Hence, proved.


x2 + xy + y2, z2 + zx + x2 and y2 + yz + z2 are in consecutive terms of an A.P.


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