Let 0

Question

$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$

$2 y^2=x z$

$\frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2}$

$x+z=4 y$

$x y+y z+z x=\frac{3}{\sqrt{2}

Vedclass · Accepted Answer

$150$

Vedclass · Answer

$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$

$2 y^2=x z$

$\frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2}$

$x+z=4 y$

$x y+y z+z x=\frac{3}{\sqrt{2}} x y z$

$y(x+z)+z x=\frac{3}{\sqrt{2}} x z \cdot y$

$4 y^2+2 y^2=\frac{3}{\sqrt{2}} y \cdot 2 y^2$

$6 y^2=3 \sqrt{2} y^3$

$y=\sqrt{2}$

$x+y+z=5 y=5 \sqrt{2}$

$3(x+y+z)^2=3 	imes 50=150$

Let $0 < z < y < x$ be three real numbers such that $\frac{1}{ x }, \frac{1}{ y }, \frac{1}{ z }$ are in an arithmetic progression and $x$, $\sqrt{2} y, z$ are in a geometric progression. If $x y+y z$ $+z x=\frac{3}{\sqrt{2}} x y z$, then $3(x+y+z)^2$ is equal to $............$.

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