Question 574789
Suppose the first three terms are a, ar, ar^2. Then we have


*[tex \LARGE a+ar+ar^2 = 14] and *[tex \LARGE a^3r^3 = 64 \Rightarrow ar = 4]


Since ar = 4, we can replace this into the first equation to get:


*[tex \LARGE \frac{4}{r} + 4 + 4r = 14]. I'll skip the algebra, but you can show that this is equivalent to 


*[tex \LARGE 2r^2 - 5r + 2 = 0]


If this quadratic has two real roots r, then we know there are two different common ratios. We can even prove this intuitively; if there are three terms {a,b,c} in a series satisfying the given conditions, then {c,b,a} must also satisfy, and the common ratio is the reciprocal of the original one.


Solving for r, we have


*[tex \LARGE r = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} = 2, \frac{1}{2}]


Since ar = 4, we can replace our known values of r to obtain a = 2 or a = 8. We have either {2,4,8} or {8,4,2}.