1. **State the problem:** We have a geometric series given by $$4(3p - 1) + 2x(3p - 1)^2 + \ldots$$ and we want to determine the values of $$x$$ for which this series converges. 2. **Identify the first term and common ratio:** The first term $$a$$ is $$4(3p - 1)$$. The second term is $$2x(3p - 1)^2$$. The common ratio $$r$$ is the ratio of the second term to the first term: $$r = \frac{2x(3p - 1)^2}{4(3p - 1)} = \frac{2x(3p - 1)^\cancel{2}}{4\cancel{(3p - 1)}} = \frac{2x(3p - 1)}{4} = \frac{x(3p - 1)}{2}$$ 3. **Convergence condition for geometric series:** A geometric series converges if and only if the absolute value of the common ratio is less than 1: $$|r| < 1$$ 4. **Apply the condition:** $$\left| \frac{x(3p - 1)}{2} \right| < 1$$ Multiply both sides by 2: $$|x(3p - 1)| < 2$$ 5. **Solve for $$x$$:** $$|x| \cdot |3p - 1| < 2$$ Divide both sides by $$|3p - 1|$$ (assuming $$3p - 1 \neq 0$$): $$|x| < \frac{2}{|3p - 1|}$$ 6. **Final answer:** The series converges for values of $$x$$ such that $$\boxed{|x| < \frac{2}{|3p - 1|}}$$ This means $$x$$ must lie strictly between $$-\frac{2}{|3p - 1|}$$ and $$\frac{2}{|3p - 1|}$$ for the series to converge.