algebra
posted by jason .
What number must be added to each of the numbers 0, 8, and 32 so that they form consecutive terms of a geometric sequence?
I don't understand what the question is asking first of all.
The answer is 4.
Help is much appreciated.

"A geometric sequence goes from one term to the next by always multiplying (or dividing) by the same value. So 1, 2, 4, 8, 16,... and 81, 27, 9, 3, 1, 1/3,... are geometric, since you multiply by 2 and divide by 3, respectively, at each step."
 http://www.purplemath.com/modules/series3.htm 
A geometric sequence cannot start with 0, or all the terms will just stay 0.
So, you want n such that each term is a constant multiple of the one before.
(8+n)/(0+n) = (32+n)/(8+n)
(8+n)^2 = n(32+n)
64 + 16n + n^2 = 32n + n^2
64 = 16n
n=4
So, the sequence starts out 4,12,36,... with each term 3x the previous one.