If 1/x+y,1/2y,1/y+z are three consecutive terms of an A.P. then prove

1.	If 1/x+y,1/2y,1/y+z are three consecutive terms of an A.P. then prove that x,y,z are three consecutive terms of a G.P.
Answer» thanks 1/x+y,1/2y,2/y+z are in A.P. 2/2y=(1/x+y)+(2/y+z) 1/y=y+z+x+y/(x+y)(y+z) 1/y=x+2y+z/xy+xz+y^2+yz xy+xz+y^2+yz=y (x+2y+z) xy+xz+y^2+yz=xy+2xz+yz xz+y^2=2xz y^2=2xz-xz y^2=xz so x,y,z are in G.P.

If 1/x+y,1/2y,1/y+z are three consecutive terms of an A.P. then prove that x,y,z are three consecutive terms of a G.P.

Answer»

thanks

1/x+y,1/2y,2/y+z are in A.P.

2/2y=(1/x+y)+(2/y+z)

1/y=y+z+x+y/(x+y)(y+z)

1/y=x+2y+z/xy+xz+y^2+yz

xy+xz+y^2+yz=y (x+2y+z)

xy+xz+y^2+yz=xy+2xz+yz

xz+y^2=2xz

y^2=2xz-xz

y^2=xz

so x,y,z are in G.P.