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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.
Solution
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.
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