Question

If a, x, b are in ap, a, Y, b are in gp and a, z, b are in hp such that x =9z and a >0, b>0 then,

(A) |y|=3z (B) x=3|y| (C) 2y=x+z (D) none of these

Answer 1

(a) |y|= 3z


x= (a+b)/2
y=√(ab)
z=(2ab)/a+b

z= (y^2)/ x
=> y^2= xz
and given x=9z
so..
9z^2 = y^2
=> 3z = |y|