If 1/y+z, 1/z+x, 1/x+y are in AP Then show that ?x^2 y^2 z?^2 in AP Sir/ Madam . which you solved is not solved in AP formula wise i feel. if so can u plz explain me thank u
Asked by | 23rd Mar, 2012, 06:34: PM
Expert Answer:
This question has being solved using the concept of AP only.
Since, 1/y+z, 1/z+x, 1/x+y are in AP, therefore the common difference will be the same. Hence, we have:
1/x+z - 1/y+z = 1/x+y - 1/z+x
So taking LCM, we get, y-x / y+z = z-y / x+y
On cross multiplying we get ,
x2 + z2 = 2y2
y2 - x2 = z2 - y2
Hence, x2, y2, z2 are in AP.
1/x+z - 1/y+z = 1/x+y - 1/z+x
So taking LCM, we get, y-x / y+z = z-y / x+y
On cross multiplying we get ,
x2 + z2 = 2y2
y2 - x2 = z2 - y2
Hence, x2, y2, z2 are in AP.
Answered by | 23rd Mar, 2012, 10:44: PM
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