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 ____________.
Answer (integer)
150
Solution
$\because \frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.
<br/><br/>$\Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{2}{y}$ ........... (i)
<br/><br/>and $x, \sqrt{2} y, z$ are in G.P.
<br/><br/>$\Rightarrow 2 y^2=x z$ .......... (ii)
<br/><br/>from (i), $\frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2}$
<br/><br/>$\Rightarrow 4 y=x+z$
<br/><br/>$$
\begin{aligned}
& \text { Also, } x y+y z+z x=\frac{3}{\sqrt{2}} x y z \\\\
& y(4 y)+x z=\frac{3}{\sqrt{2}}\left(2 y^2\right) y \\\\
& \Rightarrow 4 y^2+2 y^2=3 \sqrt{2} y^3 \\\\
& \Rightarrow 6 y^2=3 \sqrt{2} y^3 \Rightarrow y=\sqrt{2} \\\\
& \therefore 3(x+y+z)^2=3(5 y)^2=3(5 \sqrt{2})^2 \\\\
& =150
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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