1. **State the problem:** Find the 10th term of a geometric sequence where the 1st term $a_1 = -3$ and the 4th term $a_4 = 81$. 2. **Recall the formula for the $n$th term of a geometric sequence:** $$a_n = a_1 \cdot r^{n-1}$$ where $r$ is the common ratio. 3. **Use the given information to find $r$:** $$a_4 = a_1 \cdot r^{3}$$ Substitute $a_4 = 81$ and $a_1 = -3$: $$81 = -3 \cdot r^{3}$$ 4. **Solve for $r^{3}$:** $$r^{3} = \frac{81}{-3} = -27$$ 5. **Find $r$ by taking the cube root:** $$r = \sqrt[3]{-27} = -3$$ 6. **Find the 10th term $a_{10}$:** $$a_{10} = a_1 \cdot r^{9} = -3 \cdot (-3)^{9}$$ 7. **Calculate $(-3)^9$:** $$(-3)^9 = (-3)^{8} \cdot (-3) = (3^8) \cdot (-3) = 6561 \cdot (-3) = -19683$$ 8. **Calculate $a_{10}$:** $$a_{10} = -3 \cdot (-19683) = 59049$$ **Final answer:** The 10th term is $59049$, which corresponds to option d.