Exams Joint Entrance Examination Mathematics Sequences and Series Question 188 Question 188 of 519 Hard NUMERICAL
Let 0 < z < y < x 0<z<y<x 0 < z < y < x be three real numbers such that 1 x , 1 y , 1 z \dfrac1x, \dfrac1y, \dfrac1z x 1 , y 1 , z 1 are in an arithmetic progression, and x , 2 y , z x, \sqrt{2}y, z x , 2 y , z are in a geometric progression. If x y + y z + z x = 3 2 x y z xy+yz+zx = \dfrac{3}{\sqrt{2}}xyz x y + y z + z x = 2 3 x y z , then 3 ( x + y + z ) 2 3(x+y+z)^2 3 ( x + y + z ) 2 is equal to \_\_\_\_\_\_\_\_.
Answer: 150
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A.P. condition: 2 y = 1 x + 1 z = x + z x z \dfrac2y=\dfrac1x+\dfrac1z=\dfrac{x+z}{xz} y 2 = x 1 + z 1 = x z x + z . G.P. condition: 2 y 2 = x z 2y^2=xz 2 y 2 = x z . Combining: 2 y = x + z 2 y 2 ⇒ x + z = 4 y \dfrac2y=\dfrac{x+z}{2y^2} \Rightarrow x+z=4y y 2 = 2 y 2 x + z ⇒ x + z = 4 y .
x y + y z + z x = y ( x + z ) + x z = 4 y 2 + 2 y 2 = 6 y 2 . xy+yz+zx = y(x+z)+xz = 4y^2+2y^2 = 6y^2. x y + y z + z x = y ( x + z ) + x z = 4 y 2 + 2 y 2 = 6 y 2 .
Given this equals 3 2 x y z = 3 2 y ( 2 y 2 ) = 3 2 y 3 \dfrac{3}{\sqrt{2}}xyz = \dfrac{3}{\sqrt{2}}y(2y^2) = 3\sqrt{2}\,y^3 2 3 x y z = 2 3 y ( 2 y 2 ) = 3 2 y 3 : 6 y 2 = 3 2 y 3 ⇒ y = 2 6y^2=3\sqrt{2}y^3 \Rightarrow y=\sqrt{2} 6 y 2 = 3 2 y 3 ⇒ y = 2 .
\[x+y+z = 5y = 5\sqrt{2} \implies 3(x+y+z)^2 = 3(50) = 150.\] Answer the question to unlock