# Geometric sequence

#### Fernando Revilla

##### Well-known member
MHB Math Helper
I quote a question from Yahoo! Answers

Find the value of x such that the following sequence forms a geometric progression...?
x-1, 3x+4, 6x+8............so i am suppose to solve this by this rule: a,b,c then b^2=ac but im just going around in circles because i have no idea how to get an answer, my textbook says the answer is -6, but i want to know the working out....any answers appreciated!
The sequence $x-1, 3x+4, 6x+8$ forms a a geometric progression if and only if:
$$\frac{3x+4}{x-1}=\frac{6x+8}{3x+4}\text{ and } x-1\neq 0\text{ and }3x+4\neq 0$$
$$3x^2+22x+24=0\Leftrightarrow\ldots \Leftrightarrow x=-6\text{ or }x=-4/3$$
But $x=-4/3$ is not a valid solution (satisfies $3x+4=0$), so the solution is $x=-6$.