Show that x2 + xy + y2, z2 + zx + x2 and y2 + yz + z2 are in consecuti

1.	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.
Answer» 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, Since, x, 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 3xd + 3d² = 3xd + 3d² Hence, proved. x² + xy + y², z² + zx + x² and y² + yz + z² are in consecutive terms of an A.P.

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.

Answer»

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,

Since,

x, 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

3xd + 3d² = 3xd + 3d²

Hence, proved.

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