Given information:
x, y, z are in A.P.
x, y, (z-1) are in G.P.

To prove:
(y-z)^2 = x

Proof:
Let the common difference of the A.P. be d.
Then, we have:
y = x + d
z = x + 2d

Also, from the second equation, we have:
y(z-1) = x(z-1)^2

Substituting the values of y and z, we get:
(x+d)(x+d-1) = x(x+2d-1)^2

Expanding and simplifying, we get:
x^2 + d^2 - 2xd + d = x^2 + 4d^2 - 4xd + x

Simplifying further, we get:
3d^2 - 2xd + d - x = 0

This is a quadratic equation in d. Solving for d, we get:
d = (x-1)/3 or d = x/3

If d = (x-1)/3, then z = x + 2d = x + 2(x-1)/3 = (5x-2)/3
Substituting this in the equation y(z-1) = x(z-1)^2, we get:
(x+d)(x+d-1) = x(x+2d-1)^2
Simplifying, we get:
(x+(x-1)/3)((x+(x-1)/3)-1) = x(x+2(x-1)/3-1)^2
Simplifying further, we get:
16x^3 - 27x^2 + 9x - 1 = 0
This can be factored as:
(x-1)(16x^2-11x+1) = 0
Solving for x, we get:
x = 1 or x = 1/16 or x = 1/4

If d = x/3, then z = x + 2d = x + 2x/3 = 5x/3
Substituting this in the equation y(z-1) = x(z-1)^2, we get:
(x+d)(x+d-1) = x(x+2d-1)^2
Simplifying, we get:
(x+x/3)((x+x/3)-1) = x(x+2x/3-1)^2
Simplifying further, we get:
16x^3 - 27x^2 + 9x - 1 = 0
This can be factored as:
(x-1)(16x^2-11x+1) = 0
Solving for x, we get:
x = 1 or x = 1/16 or x = 1/4

Therefore, the only possible value of x is 1/4.

Substituting x = 1/4 in the equations for y and z, we get:
y = 1/4 + d = 1/4 + (1/4)/3 = 7/12
z = 1/4 + 2d = 1/4 + 2(1/4)/3 = 5/6

Therefore, the A.P.