Do you mean
1/(y-x), 1/(2y), and 1/(y-z)
or
1/(y-x), (1/2)y, and (1/y)-z
??
Given that 1/(y-x), 1/2y, and 1/y-z are consecutive terms of an arithmetic progression, prove that x,y, and z are consecutive terms of a geometric progression.
4 answers
1/(y-x), 1/(2y), 1/(y-z)
well, just plug and chug
1/(2y) - 1/(y-x) = 1/(y-z) - 1/(2y)
(x+y)/(2xy-2y^2) = (y+z)/(2y^2-2yz)
(x+y)/(x-y) = (y+z)/(y-z)
(x+y)(y-z) = (y+z)(x-y)
xy+y^2-xz-yz = xy+xz-y^2-yz
y^2-xz = xz-y^2
y^2 = xz
y/x = z/y
Thus x,y,z form a geometric progression
1/(2y) - 1/(y-x) = 1/(y-z) - 1/(2y)
(x+y)/(2xy-2y^2) = (y+z)/(2y^2-2yz)
(x+y)/(x-y) = (y+z)/(y-z)
(x+y)(y-z) = (y+z)(x-y)
xy+y^2-xz-yz = xy+xz-y^2-yz
y^2-xz = xz-y^2
y^2 = xz
y/x = z/y
Thus x,y,z form a geometric progression
How did you get this equation (x+y)/(2xy-2y^2) = (y+z)/(2y^2-2yz) ?
Can you show your solution?
Can you show your solution?