Let $0<\mathrm{z}<\mathrm{y}<\mathrm{x}$ be three real numbers such that $\frac{1}{\mathrm{x}}, \frac{1}{\mathrm{y}}, \frac{1}{\mathrm{z}}$ 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 ______.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Sequences 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

2 y

z

are in a geometric progression. If

x y + y z + z x = \frac{3}{2} x y z

, then

3 (x + y + z)^{2}

is equal to ______.

Answer: 150

From AP condition:

\frac{2}{y} = \frac{1}{x} + \frac{1}{z}

... (i). From GP condition:

(2 y)^{2} = x z

, so

2 y^{2} = x z

... (ii). From the given relation:

\frac{x y + y z + z x}{x y z} = \frac{3}{2}

, which gives

\frac{1}{z} + \frac{1}{x} + \frac{1}{y} = \frac{3}{2}

... (iii). From (i) and (iii):

\frac{2}{y} + \frac{1}{y} = \frac{3}{2}

, so

\frac{3}{y} = \frac{3}{2}

, giving

y = 2

. From (ii):

x z = 2 (2)^{2} = 4

. From (i):

x + z = \frac{2 x y}{y} = 22 \cdot 2 = 42

(using

\frac{1}{x} + \frac{1}{z} = \frac{2}{2} = 2

). Solving

x + z = 42

and

x z = 4

x = 2 (2 + 1)

z = 2 (2 - 1)

. Thus

x + y + z = 42 + 2 = 52

, and

3 (x + y + z)^{2} = 3 (52)^{2} = 3 (50) = 150

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Integer answer