## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$x=-4$
A sequence is geometric if there exists a common ratio $r$ among consecutive terms such as: $r=\dfrac{a_n}{a_{n−1}}$. In order for a sequence to be geometric, the quotient of all consecutive terms must remain constant. So, we have: $\dfrac{a_3}{a_2}=\dfrac{a_2}{a_1}$. Substituting the given values of the first three terms, we obtain: $$\dfrac{x+3}{x+2}=\dfrac{x+2}{x}$$ Next, we will do cross-multiplication to obtain: \begin{align*}\dfrac{x+3}{x+2}&=\dfrac{x+2}{x}\\x(x+3)&=(x+2)(x+2)\\ x^2+3x&=x^2+4x+4\\4x+4&=3x\\ 4x+4-3x&=3x-3x\\ x+4&=0\\ x&=-4\end{align*}