Answer:
Given,
A1 = AM of x and y
A2 = AM of y and z
A1 = (x+y)/2
A2 = (y+x)/2
AM of A1 and A2 = (A1 + A2)/2
=> [(x+y)/2 + (y+z)/2]/2
=> [x+y+y+z]/2
=> [x+2y+z]/2
x, y, z are in AP, y = (x+z)/2
AM = [(x + z/2) + (2y/2)]/2
AM = (y + y)/2
AM = 2y/2
AM = y
Thus, proved.