1. **State the problem:** Determine if the sequence $7, -\frac{7}{3}, \frac{7}{9}, \frac{7}{27}, \ldots$ is geometric. If it is, find the common ratio $r$. 2. **Recall the definition:** A sequence is geometric if the ratio between consecutive terms is constant. This ratio is called the common ratio $r$. 3. **Calculate the ratio between the second and first term:** $$r = \frac{-\frac{7}{3}}{7} = -\frac{7}{3} \times \frac{1}{7} = -\frac{\cancel{7}}{3} \times \frac{1}{\cancel{7}} = -\frac{1}{3}$$ 4. **Calculate the ratio between the third and second term:** $$r = \frac{\frac{7}{9}}{-\frac{7}{3}} = \frac{7}{9} \times \frac{-3}{7} = \frac{\cancel{7}}{9} \times \frac{-3}{\cancel{7}} = -\frac{3}{9} = -\frac{1}{3}$$ 5. **Calculate the ratio between the fourth and third term:** $$r = \frac{\frac{7}{27}}{\frac{7}{9}} = \frac{7}{27} \times \frac{9}{7} = \frac{\cancel{7}}{27} \times \frac{9}{\cancel{7}} = \frac{9}{27} = \frac{1}{3}$$ 6. **Check consistency:** The first two ratios are $-\frac{1}{3}$ but the third ratio is $\frac{1}{3}$. Since the ratios are not all equal, the sequence is **not geometric**. **Final answer:** The sequence is not geometric.