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Question
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.
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Solution
\[\text {As, x, y and z are in A . P } . \]
\[\text { So, y } = \frac{x + z}{2} . . . . . \left( i \right)\]
\[\text { Now }, \]
\[\left( x^2 + xy + y^2 \right) + \left( y^2 + yz + z^2 \right)\]
\[ = x^2 + z^2 + 2 y^2 + xy + yz\]
\[ = x^2 + z^2 + 2 y^2 + y\left( x + z \right)\]
\[ = x^2 + z^2 + 2 \left( \frac{x + z}{2} \right)^2 + \left( \frac{x + z}{2} \right)\left( x + z \right) \left[ \text { Using } \left( i \right) \right]\]
\[ = x^2 + z^2 + 2\left( \frac{\left( x + z \right)^2}{4} \right) + \frac{\left( x + z \right)^2}{2}\]
\[ = x^2 + z^2 + \frac{\left( x + z \right)^2}{2} + \frac{\left( x + z \right)^2}{2}\]
\[ = x^2 + z^2 + \left( x + z \right)^2 \]
\[ = x^2 + z^2 + x^2 + 2xy + z^2 \]
\[ = 2 x^2 + 2xy + 2 z^2 \]
\[ = 2\left( x^2 + xy + z^2 \right)\]
\[\text { Since, } \left( x^2 + xy + y^2 \right) + \left( y^2 + yz + z^2 \right) = 2\left( x^2 + xy + z^2 \right)\]
\[\text { So, } \left( x^2 + xy + y^2 \right), \left( x^2 + xy + z^2 \right) \text { and } \left( y^2 + yz + z^2 \right) \text { are in A . P } .\]
Hence, x2 + xy + y2, z2 + zx + x2 and y2 + yz + z2 are consecutive terms of an A.P.
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