1. The problem asks to find a recursive formula for the sequence defined explicitly by $$a_n = 10(-4)^{n-1}$$ that depends on the previous term $$a_{n-1}$$. 2. Recall that a recursive formula expresses $$a_n$$ in terms of $$a_{n-1}$$. We start by writing $$a_n$$ and $$a_{n-1}$$ explicitly: $$a_n = 10(-4)^{n-1}$$ $$a_{n-1} = 10(-4)^{(n-1)-1} = 10(-4)^{n-2}$$ 3. To find the recursive relation, divide $$a_n$$ by $$a_{n-1}$$: $$\frac{a_n}{a_{n-1}} = \frac{10(-4)^{n-1}}{10(-4)^{n-2}} = \cancel{10} \cdot (-4)^{n-1} \cdot \frac{1}{\cancel{10} (-4)^{n-2}} = (-4)^{(n-1)-(n-2)} = (-4)^1 = -4$$ 4. This shows that each term is $$-4$$ times the previous term. Therefore, the recursive formula is: $$a_n = -4 a_{n-1}$$ 5. The initial term is given by substituting $$n=1$$ in the explicit formula: $$a_1 = 10(-4)^{0} = 10 \cdot 1 = 10$$ 6. Final recursive formula with initial condition: $$\boxed{a_1 = 10, \quad a_n = -4 a_{n-1} \text{ for } n \geq 2}$$