Question -
Answer -
x, y, z are in AP
Given, 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 = commondifference,
So, Y = x + d and x =x + 2d
Let us consider theLHS:
(z2 +zx + x2) – (x2 + xy + y2)
z2 +zx – xy – y2
(x + 2d)2 +(x + 2d)x – x(x + d) – (x + d)2
x2 +4xd + 4d2 + x2 + 2xd – x2 – xd– x2 – 2xd – d2
3xd + 3d2
Now, let us considerRHS:
(y2 +yz + z2) – (z2 + zx + x2)
y2 +yz – zx – x2
(x + d)2 +(x + d)(x + 2d) – (x + 2d)x – x2
x2 +2dx + d2 + x2 + 2dx + xd + 2d2 –x2 – 2dx – x2
3xd + 3d2
LHS = RHS
∴ x2 +xy + y2, z2 + zx + x2 and y2 +yz + z2 are in consecutive terms of A.P
Hence proved.