Let 0 < z < y < x be three real numbers such that 1 x , 1 y , 1 z are in an arithmetic progression and x , 2 y , z are in a geometric progression. If x y + y z + z x = 3 2 x y z , then 3 ( x + y + z ) 2 is equal to

Step-by-step solution

Given that 1 x , 1 y , 1 z are in AP and x , 2 y , z are in GP. As given, 2 y = 1 x + 1 y   . . . . . . . i Also, 2 y 2 = x z   . . . . . . . ii Also given that x y + y z + z x = 3 2 x y z ⇒ 1 x + 1 y + 1 z = 3 2   . . . . . iii From ( i ) and ( i i i ) we get 3 y = 3 2 y = 2   . . . . . iv Now from ( i i ) x z = 4     . . . . . . . v Now using ( i i ) ,   ( i v )   and   ( v ) ⇒ x + z = 4 2 Hence 3 ( x + y + z ) 2 = 3 ( 2 + 4 2 ) 2 = 150 Therefore, this is t