Show that $x^{2} + x y + y^{2}, z^{2} + z x + x^{2}$ and $y^{2} + y z + z^{2}$ are consecutive terms of an A.P., if $x, y$ and $z$ are in A.P.

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Solution

The terms $(x^{2} + x y + y^{2}), (z^{2} + x z + x^{2})$ and $y^{2} + y z + z^{2}$ will be in A.P. if
$(z^{2} + x z + x^{2}) - (x^{2}) - (x^{2} + x y + y^{2}) = (y^{2} + y z + z^{2}) - (z^{2} + x z + x^{2})$
i.e., $z^{2} + x z - x y - y^{2} = y^{2} + y z - x z - x^{2}$
i.e., $x^{2} + z^{2} + 2 x z - y^{2} = y^{2} + y z + x y$
i.e., $(x + z)^{2} - y^{2} = y (x + y + z)$
i.e., $x + z - y = y$
i.e., $x + z = 2 y$
Which is true, since $x, y, z$ are in A.P.
Hence $x^{2} + x y + y^{2}, z^{2} + x z + x^{2}, y^{2} + y z + z^{2}$ are in A.P.

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