Find x such that x - 4, x, and 3x - 8 are three consecutive terms in a geometric sequence

In a geometric sequence dividing any term by the term before it yields the common multiplier.

Then x / (x+8) = (3x-8) / x

Solving for x: reduces to x^2-10x + 16 factor into (x-2) (x-8)

Then x = 2 or 8

2 produces a sequence of - 2, 2, - 2 ...

8 produces a sequence of 4, 8, 16 ...