Question

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

Answer 1

The terms (x² + xy + y²), (z² + xz + x²) and (y² + yz + z²) will be in A.P.

if (z² + xz + x²) – (x² + xy + y²) = (y² + yz + z²) – (z² + xz + x²)

i.e., z² + xz – xy – y² = y² + yz – xz – x²

i.e., x² + z² + 2xz – y² = y² + yz + xy

i.e., (x + z)² – y² = y(x + y + z)

i.e., x + z – y = y

i.e., x + z = 2y

Which is true, since x, y, z are in A.P.

Hence x² + xy + y² , z² + xz + x², y² + yz + z² are in A.P.