1. **State the problem:** We have a sequence starting with terms -12, -9, -6, ... and we need to find the value of $x$ given that the ratio of the 8th term to the 6th term is $\frac{1}{3x}$. 2. **Identify the sequence type:** The sequence is arithmetic because the difference between consecutive terms is constant. The common difference $d$ is calculated as: $$d = -9 - (-12) = 3$$ 3. **General term formula for arithmetic sequence:** The $n$th term $a_n$ is given by $$a_n = a_1 + (n-1)d$$ where $a_1 = -12$ and $d = 3$. 4. **Find the 6th and 8th terms:** $$a_6 = -12 + (6-1) \times 3 = -12 + 15 = 3$$ $$a_8 = -12 + (8-1) \times 3 = -12 + 21 = 9$$ 5. **Set up the ratio and solve for $x$:** Given $$\frac{a_8}{a_6} = \frac{1}{3x}$$ Substitute values: $$\frac{9}{3} = \frac{1}{3x}$$ Simplify left side: $$3 = \frac{1}{3x}$$ Cross multiply: $$3 \times 3x = 1$$ $$9x = 1$$ $$x = \frac{1}{9}$$ **Final answer:** $$x = \frac{1}{9}$$