x^2,y^2,z^2 are in A.P. then (x+y),(y+z)and(x+z) are in ---?

Asked by nivedithawarrier | 2nd Nov, 2010, 09:55: PM

Expert Answer:

Dear Student,
 
As x2, y2 and z2 are in AP, so,
2y2 = x2 + z2.........................................(1)
Now, if we consider the given terms, (x+y),(y+z)and(x+z); as AP, we will get:
2(y+z) = (x+y) + (x+z)
=> y + z = 2x
which does not relate to equation (1)
Similarly if we consider them to be in GP, we will get:
(y+z)2 = (x+y)(x+z)
=>y2 + z2 +yz = x2 +xz + xy
which, again, does not relate to equation (1)
and finally on considering them to be in HP:
2/(y+z) = 1/(x+y) + 1/(x+Z)
=> 2x2 = y2 + z2
which says y2, x2, z2 are in AP, which is not true!
 
So, there is no specific series that the terms (x+y),(y+z)and(x+z) follow.
 
Regards Topperlearning.

Answered by  | 5th Nov, 2010, 12:02: PM

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