Q.

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=32xyz, then 3(x+y+z)2 is equal to _________ .         [2023]


Ans.

(150)

2y=1x+1z  and  2y2=xz

⇒2y=x+zxz=x+z2y2⇒x+z=4y

Now, xy+yz+zx=32xyz⇒y(x+z)+zx=32xz·y

⇒4y2+2y2=32y·2y2⇒6y2=32y3⇒y=2

∴  x+y+z=5y=52⇒3(x+y+z)2=3×50=150