Question

If 1/x+y, 1/2y,1/y+z are in A.P. Prove that x,y,z are in G.P.

Answer 1

Hey!!!!....here is ur answer

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.

Hope it will help you ☺

Answer 2

Answer:

Step-by-step explanation:

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.