1. **Stating the problem:** We want to find the nth term of the sequence 3, 6, 7, 9. 2. **Observing the sequence:** The terms are 3, 6, 7, 9. Let's look at the differences between consecutive terms: - $6 - 3 = 3$ - $7 - 6 = 1$ - $9 - 7 = 2$ 3. **Analyzing the pattern:** The differences are 3, 1, 2, which do not form a simple arithmetic or geometric progression. 4. **Attempting to find a formula:** Since the differences are irregular, let's check if the sequence can be expressed as a piecewise or recursive formula, or if it repeats a pattern. 5. **Checking for repetition:** The differences 3,1,2 might repeat. If so, the next difference would be 3 again, and the sequence would continue as: - $9 + 3 = 12$ - $12 + 1 = 13$ - $13 + 2 = 15$ 6. **Generalizing the nth term:** The differences cycle every 3 terms: 3,1,2. 7. **Define the difference function:** $$d(n) = \begin{cases} 3 & \text{if } n \equiv 1 \pmod{3} \\ 1 & \text{if } n \equiv 2 \pmod{3} \\ 2 & \text{if } n \equiv 0 \pmod{3} \end{cases}$$ 8. **Express the nth term:** Starting from $a_1 = 3$, the nth term is: $$a_n = 3 + \sum_{k=1}^{n-1} d(k)$$ 9. **Calculating the sum of differences:** Since the differences repeat every 3 terms with sum $3 + 1 + 2 = 6$, for $m = \lfloor \frac{n-1}{3} \rfloor$ full cycles and remainder $r = (n-1) \bmod 3$, we have: $$\sum_{k=1}^{n-1} d(k) = 6m + \sum_{k=1}^r d(k)$$ 10. **Final formula:** $$a_n = 3 + 6m + \sum_{k=1}^r d(k)$$ where $m = \left\lfloor \frac{n-1}{3} \right\rfloor$ and $r = (n-1) \bmod 3$. This formula allows you to find the nth term of the sequence by summing the repeating differences.