Let 0<z<y<x be three real numbers such that 1x, 1y, 1z are in an arithmetic progression and x, 2y,z are in a geometric progression. If xy+yz+zx= 3 2 xyz, then 3(x+y+z)^2 is equal to

JEE Main Mathematics question

Let be three real numbers such that are in an arithmetic progression and are in a geometric progression. If xy+yz+zx=\frac3{\sqrt2}xyz, then is equal to

Answer: 150

Solution

Dividing xy+yz+zx=\frac3{\sqrt2}xyz by gives , and the AP condition turns it into , so

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Question ID VK-022175