JEE Mains · Maths · STD 11 - 8. sequence and series
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 \(............\).
- A \(150\)
- B \(140\)
- C \(130\)
- D \(120\)
Answer & Solution
Correct Answer
(A) \(150\)
Step-by-step Solution
Detailed explanation
\(\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\) \(2 y^2=x z\) \(\frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2}\) \(x+z=4 y\) \(x y+y z+z x=\frac{3}{\sqrt{2}} x y z\) \(y(x+z)+z x=\frac{3}{\sqrt{2}} x z \cdot y\) \(4 y^2+2 y^2=\frac{3}{\sqrt{2}} y \cdot 2 y^2\) \(6 y^2=3 \sqrt{2} y^3\)…
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