Find all possible values of x when first 3 terms of a geometric progression are: x-1, 2-x and x/3

Yep the question's in the title :P
How would i go about finding the values of x? It's not so much the answer i'm interested in more how to get there.

Hello,

In a geometric progression, any two successive terms are linked by the equality:
U(n+1) = q.U(n) where q is the reason of the progression.

If x-1, 2-x and x/3 are the first terms of a geometric progression, we would then have:
x/3 = q(2 - x)
and
2 - x = q(x - 1)

If x=1, the first three terms would be 0, 2 and ⅓, which is obviously not a geometric progression. So x is not equal to 1.
If x=2, the first three terms would be 1, 0 and ⅔, which is obviously not a geometric progression. So x is not equal to 2.
Since x≠1 and x≠2, we can divide by x-1 and 2-x:

x/3 / (2 - x) = q = (2 - x)/(x - 1)
(2 - x)² = (x - 1)x/3
4 - 4x + x² = (x² - x)/3
12 - 12x + 3x² = x² - x
2x² - 11x + 12 = 0
2x³ - 3x - 8x + 12 = 0
x(2x - 3) - 4(2x - 3) = 0
(2x - 3)(x - 4) = 0
x = 3/2 or x = 4

Check if x=4:

x - 1 = 4 - 1 = 3
2 - x = 2 - 4 = -2
x / 3 = 4/3

(2 - x) / (x - 1) = -⅔
(x/3) / (2 - x) = (4/3) / (-2) = -⅔
QED.

Check if x=3/2:

x - 1 = 3/2 - 1 = ½
2 - x = 2 - 3/2 = ½
x / 3 = (3/2)/3 = ½

QED

So there are 2 values of x that would make x-1, 2-x an x/3 the three first terms of an arithmetic progression. And those values are x = 4 and x = 3/2.

Logically,
Dragon.Jade :-)

(2 - x)/(x - 1) = (x/3)/(2 - x)

=> (2 - x)^2 = (x/3)(x - 1)

=> 3(4 - 4x + x^2) = x^2 - x

i.e. 12 - 12x + 3x^2 = x^2 - x

=> 2x^2 - 11x + 12 = 0

so, (2x - 3)(x - 4) = 0

=> x = 3/2 or 4

:)>

(2nd term)/(first term)=(3rd term)/(2nd term)

(2-x)/(x-1)=(x/3)/(2-x)