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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

The terms (x2+xy+y2),(z2+xz+x2) and y2+yz+z2 will be in A.P. if
(z2+xz+x2)(x2)(x2+xy+y2)=(y2+yz+z2)(z2+xz+x2)
i.e., z2+xzxyy2=y2+yzxzx2
i.e., x2+z2+2xzy2=y2+yz+xy
i.e., (x+z)2y2=y(x+y+z)
i.e., x+zy=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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