Let 0<z<y<x be three real numbers such that 1 x, 1 y, 1 z are in an arithmetic progression and x,√2y,z are in a geometric progression. If xy+yz+zx= 3…
JEE Main 2023 — Mathematics Algebra
2023integermedium
Let 0<z<y<x be three real numbers such that x1,y1,z1 are in an arithmetic progression and x,2y,z are in a geometric progression. If xy+yz+zx=23xyz, then 3(x+y+z)2 is equal to
Official previous-year question
Held on 8 Apr 2023 · Verified 6 Jul 2026.
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Solution
Given that x1,y1,z1 are in AP and x,2y,z are in GP.
As given, y2=x1+y1.......(i)
Also, 2y2=xz.......(ii)
Also given that xy+yz+zx=23xyz
⇒x1+y1+z1=23.....(iii)
From (i) and (iii) we get y3=23
y=2.....(iv)
Now from (ii)xz=4.......(v)
Now using (ii),(iv)and(v)
⇒x+z=42
Hence 3(x+y+z)2=3(2+42)2
=150
Therefore, this is the required answer.
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