Let be three real numbers such that are in an arithmetic progression and ,are in a geometric progression. If xyz, then 3 (x+y+z) 2 =________.

150 Explanation: \begin{aligned} & \because\frac{1}{x},\frac{1}{y},\frac{1}{z}\text{are in A.P.} \\ & \Rightarrow\frac{1}{x}+\frac{1}{z}=\frac{2}{y} \\ & \mathrm{and}x,\sqrt{2}y,z\text{are in G.P.} \\ & \Rightarrow2y^{2}=xz \\ & \mathrm{from~(i),~}\frac{2}{y}=\frac{x+z}{xz}=\frac{x+z}{2y^{2}} \\ & \Rightarrow4y=x+z \\ & \mathrm{Also,}xy+yz+zx=\frac{3}{\sqrt{2}}xyz \\ & y(4y)+xz=\frac{3}{\sqrt{2}}(2y^{2})y \\ & \Rightarrow4y^2+2y^2=3\sqrt{2}y^3 \\ & \Rightarrow6y^{2}=3\sqrt{2}y^{3}\Rightarrow y=\sqrt{2} \\ & \therefore3(x+y+z)^{2}=3(5y)^{2}=3(5\sqrt{2})^{2} \\ & =150 \end{aligned}

How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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 $x, 2$ ,are in a geometric progression. If $x y + y z + z x = 3/ 2$ xyz, then 3 (x+y+z)²=________.

Choose the correct answer: