1. The problem states that the sequence starts at 14 and increases by 8 each time. This is an arithmetic sequence with first term $a_1=14$ and common difference $d=8$. 2. The $n$th term of an arithmetic sequence is given by the formula: $$a_n = a_1 + (n-1)d$$ 3. Substitute the known values: $$a_n = 14 + (n-1) \times 8 = 14 + 8n - 8 = 8n + 6$$ 4. We want to find the term closest to 100. Set $a_n$ close to 100: $$8n + 6 \approx 100$$ 5. Solve for $n$: $$8n = 100 - 6 = 94$$ $$n = \frac{94}{8} = 11.75$$ 6. Since $n$ must be an integer, check terms for $n=11$ and $n=12$: - For $n=11$: $$a_{11} = 8 \times 11 + 6 = 88 + 6 = 94$$ - For $n=12$: $$a_{12} = 8 \times 12 + 6 = 96 + 6 = 102$$ 7. Compare which is closer to 100: - $|94 - 100| = 6$ - $|102 - 100| = 2$ 8. The term $a_{12} = 102$ is closer to 100. **Final answer:** The 12th term, 102, is the number in the sequence closest to 100.