are consecutive terms of an A.P., if x, y and z are in A.P.

Solution:

As, $x, y$ and $z$ are in A.P.

So, $y=\frac{x+z}{2} \quad \ldots \ldots($ i)

Now,

$\left(x^{2}+x y+y^{2}\right)+\left(y^{2}+y z+z^{2}\right)$

$=x^{2}+z^{2}+2 y^{2}+x y+y z$

$=x^{2}+z^{2}+2 y^{2}+y(x+z)$

$=x^{2}+z^{2}+2\left(\frac{x+z}{2}\right)^{2}+\left(\frac{x+z}{2}\right)(x+z) \quad[$ Using $(\mathrm{i})]$

$=x^{2}+z^{2}+2\left(\frac{(x+z)^{2}}{4}\right)+\frac{(x+z)^{2}}{2}$

$=x^{2}+z^{2}+\frac{(x+z)^{2}}{2}+\frac{(x+z)^{2}}{2}$

$=x^{2}+z^{2}+(x+z)^{2}$

$=x^{2}+z^{2}+x^{2}+2 x y+z^{2}$

$=2 x^{2}+2 x y+2 z^{2}$

$=2\left(x^{2}+x y+z^{2}\right)$

Since, $\left(x^{2}+x y+y^{2}\right)+\left(y^{2}+y z+z^{2}\right)=2\left(x^{2}+x y+z^{2}\right)$

So, $\left(x^{2}+x y+y^{2}\right),\left(x^{2}+x y+z^{2}\right)$ and $\left(y^{2}+y z+z^{2}\right)$ are in A. P.

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