1. **State the problem:** We have a geometric sequence with the first term $a_1 = -2$ and common ratio $r = -\frac{1}{4}$. We need to find the next three terms $a_2$, $a_3$, and $a_4$. 2. **Formula for the $n$th term of a geometric sequence:** $$a_n = a_1 \times r^{n-1}$$ This means each term is the first term multiplied by the common ratio raised to the power of one less than the term number. 3. **Calculate the next three terms:** - For $a_2$: $$a_2 = -2 \times \left(-\frac{1}{4}\right)^{2-1} = -2 \times \left(-\frac{1}{4}\right) = \frac{2}{4} = \frac{1}{2}$$ - For $a_3$: $$a_3 = -2 \times \left(-\frac{1}{4}\right)^{3-1} = -2 \times \left(-\frac{1}{4}\right)^2 = -2 \times \frac{1}{16} = -\frac{2}{16} = -\frac{1}{8}$$ - For $a_4$: $$a_4 = -2 \times \left(-\frac{1}{4}\right)^{4-1} = -2 \times \left(-\frac{1}{4}\right)^3 = -2 \times \left(-\frac{1}{64}\right) = \frac{2}{64} = \frac{1}{32}$$ 4. **Final answer:** The next three terms of the sequence are $\frac{1}{2}$, $-\frac{1}{8}$, and $\frac{1}{32}$.