Show that x² + xy + y², z² + zx + x² and y² + yz + z² are consecutive terms of an A.P., if x, y and z are in A.P. [NCERT EXEMPLAR]

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Solution

$As, x, y and z are in A . P . So, y = \frac{x + z}{2} . . . . . (i) Now, (x^{2} + x y + y^{2}) + (y^{2} + y z + z^{2}) = x^{2} + z^{2} + 2 y^{2} + x y + y z = x^{2} + z^{2} + 2 y^{2} + y (x + z) = x^{2} + z^{2} + 2 {(\frac{x + z}{2})}^{2} + (\frac{x + z}{2}) (x + z) [Using (i)] = x^{2} + z^{2} + 2 (\frac{{(x + z)}^{2}}{4}) + \frac{{(x + z)}^{2}}{2} = x^{2} + z^{2} + \frac{{(x + z)}^{2}}{2} + \frac{{(x + z)}^{2}}{2} = x^{2} + z^{2} + {(x + z)}^{2} = x^{2} + z^{2} + x^{2} + 2 x y + z^{2} = 2 x^{2} + 2 x y + 2 z^{2} = 2 (x^{2} + x y + z^{2}) Since, (x^{2} + x y + y^{2}) + (y^{2} + y z + z^{2}) = 2 (x^{2} + x y + z^{2}) So, (x^{2} + x y + y^{2}), (x^{2} + x y + z^{2}) and (y^{2} + y z + z^{2}) are in A . P .$

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

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