If x2 (y + z), y2 (z + x), z2(x + y) are in A.P., then prove that ei

1.	If x2 (y + z), y2 (z + x), z2(x + y) are in A.P., then prove that either x, y, z are in A.P. or xy + yz + zx = 0.
Answer» ∵ x²(y + z), y²(z + x), z²(x + y) are in A.P. ∴ Adding xyz in each terms x²(y + z) + xyz, y²(z + x) + xyz, z²(x + y) + xyz also will be in A.P. or x(xy + yz + zx), y(xy +yz + zx), z(xy + yz + zx) also will be in A.P. ∴ 2y(xy + yz + zx) = x(xy + yz + zx) + z(xy + yz + zx) ⇒ 2y(xy + yz + zx) = (xy + yz + zx) (x + z) ⇒ 2y(xy + yz + zx) – (xy + yz + zx) (x + z) = 0 ⇒ (xy + yz + zx) (2y – x – z) = 0 If 2y – x – z = 0 Then 2y = x + z ⇒ x, y, z are in A.P. or xy + yz + zx = 0 Hence Proved.

If x2 (y + z), y2 (z + x), z2(x + y) are in A.P., then prove that either x, y, z are in A.P. or xy + yz + zx = 0.

Answer»

∵ x²(y + z), y²(z + x), z²(x + y) are in A.P.

∴ Adding xyz in each terms x²(y + z) + xyz, y²(z + x) + xyz, z²(x + y) + xyz also will be in A.P.

or x(xy + yz + zx), y(xy +yz + zx), z(xy + yz + zx)

also will be in A.P.

∴ 2y(xy + yz + zx) = x(xy + yz + zx) + z(xy + yz + zx)

⇒ 2y(xy + yz + zx) = (xy + yz + zx) (x + z)

⇒ 2y(xy + yz + zx) – (xy + yz + zx) (x + z) = 0

⇒ (xy + yz + zx) (2y – x – z) = 0

If 2y – x – z = 0

Then 2y = x + z

⇒ x, y, z are in A.P.

or xy + yz + zx = 0

Hence Proved.