1. **State the problem:** You start with the number 5036. Then you add 60, subtract 462, and divide the result by 1.5. You want to find the 10th number in this sequence. 2. **Understand the sequence:** Each step applies the same operations: add 60, subtract 462, then divide by 1.5. 3. **Write the formula for the next term:** If $a_n$ is the $n$th term, then $$a_{n+1} = \frac{a_n + 60 - 462}{1.5} = \frac{a_n - 402}{1.5}$$ 4. **Simplify the formula:** $$a_{n+1} = \frac{a_n - 402}{1.5} = \frac{a_n - 402}{\frac{3}{2}} = \frac{2}{3}(a_n - 402)$$ 5. **Calculate terms step-by-step:** - Start: $a_1 = 5036$ - Calculate $a_2$: $$a_2 = \frac{2}{3}(5036 - 402) = \frac{2}{3} \times 4634 = \frac{2 \times 4634}{3} = \frac{9268}{3} \approx 3089.33$$ 6. **Calculate $a_3$:** $$a_3 = \frac{2}{3}(a_2 - 402) = \frac{2}{3}\left(\frac{9268}{3} - 402\right) = \frac{2}{3}\left(\frac{9268}{3} - \frac{1206}{3}\right) = \frac{2}{3} \times \frac{8062}{3} = \frac{2 \times 8062}{9} = \frac{16124}{9} \approx 1791.56$$ 7. **Recognize the pattern:** This is a linear recurrence relation. The general solution is $$a_n = A \left(\frac{2}{3}\right)^{n-1} + L$$ where $L$ is the fixed point satisfying $L = \frac{2}{3}(L - 402)$. 8. **Find the fixed point $L$:** $$L = \frac{2}{3}(L - 402) \Rightarrow 3L = 2L - 804 \Rightarrow 3L - 2L = -804 \Rightarrow L = -804$$ 9. **Find constant $A$ using initial condition:** $$a_1 = A + L = 5036 \Rightarrow A = 5036 + 804 = 5840$$ 10. **Write the explicit formula:** $$a_n = 5840 \left(\frac{2}{3}\right)^{n-1} - 804$$ 11. **Calculate the 10th term:** $$a_{10} = 5840 \left(\frac{2}{3}\right)^9 - 804$$ Calculate $\left(\frac{2}{3}\right)^9$: $$\left(\frac{2}{3}\right)^9 = \frac{2^9}{3^9} = \frac{512}{19683} \approx 0.026$$ So, $$a_{10} \approx 5840 \times 0.026 - 804 = 151.84 - 804 = -652.16$$ **Final answer:** The 10th number in the sequence is approximately $-652.16$.