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.

How did you get this equation (x+y)/(2xy-2y^2) = (y+z)/(2y^2-2yz) ?

Can you show your solution?

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

To prove that x, y, and z are consecutive terms of a geometric progression, we need to show that the ratios between them are constant.

Let's start by finding the ratios between the given terms:

1/(y - x), 1/2y, 1/(y - z)

The formula for an arithmetic progression is given by:

An = A1 + (n - 1)d

Where An is the nth term, A1 is the first term, n is the position of the term in the sequence, and d is the common difference. In this case, we have:

1/(y - x) = A1 + (n - 1)d1 ... (1)
1/2y = A1 + (n - 1)d2 ... (2)
1/(y - z) = A1 + (n - 1)d3 ... (3)

From equation (2), we can rearrange it as follows:

1/2y = A1 + (n - 1)d2
=> (n - 1)d2 = 1/2y - A1
=> d2 = (1/2y - A1)/(n - 1)

Similarly, from equations (1) and (3), we have:

d1 = (1/(y - x) - A1)/(n - 1)
d3 = (1/(y - z) - A1)/(n - 1)

Now, to prove that x, y, and z are in a geometric progression, we need to show that the ratios d2/d1 and d3/d2 are constant.

d2/d1 = [(1/2y - A1)/(n - 1)] / [(1/(y - x) - A1)/(n - 1)]
= (1/2y - A1) / (1/(y - x) - A1)
= (y - x)/(2y - (y - x))
= (y - x)/(y + x)

Similarly, d3/d2 can be found as:

d3/d2 = [(1/(y - z) - A1)/(n - 1)] / [(1/2y - A1)/(n - 1)]
= (1/(y - z) - A1) / (1/2y - A1)
= (2y - z)/(y - z)

Since d2/d1 = (y - x)/(y + x) and d3/d2 = (2y - z)/(y - z), we can see that these ratios are constant.

Therefore, we have proved that x, y, and z are consecutive terms of a geometric progression.

1/(y-x), 1/(2y), 1/(y-z)

Do you mean

1/(y-x), 1/(2y), and 1/(y-z)
or
1/(y-x), (1/2)y, and (1/y)-z

??