Question:

<p><strong>Step 1:</strong> Use the A.P. sum formula: $S_n = \frac{n}{2}(2a + (n-1)d)$</p><p><strong>Step 2:</strong> Note that $S_{3n} - S_{n-1}$ represents the sum of terms from position $n$ to position $3n$, which contains $2n+1$ terms.</p><p><strong>Step 3:</strong> Use the key identity: $S_{3n} - S_{n-1} = a_n + a_{n+1} + \cdots + a_{3n}$ (sum of $2n+1$ consecutive A.P. terms)</p><p>This sum = $\frac{2n+1}{2}(a_n + a_{3n}) = \frac{2n+1}{2}(2a + (n-1)d + 2a + (3n-1)d) = (2n+1)(2a + (2n-1)d)$</p><p><strong>Step 4:</strong> Similarly, $S_{2n} - S_{2n-1} = a_{2n}$ (the $2n$-th term) = $a + (2n-1)d$</p><p><strong>Step 5:</strong> Set up the ratio:</p><p>$$\frac{(2n+1)(2a + (2n-1)d)}{a + (2n-1)d} = 31$$</p><p><strong>Step 6:</strong> Let $b = a + (2n-1)d$. Then $2a + (2n-1)d = a + b$</p><p>$$\frac{(2n+1)(a + b)}{b} = 31$$</p><p>$$(2n+1)\left(1 + \frac{a}{b}\right) = 31$$</p><p><strong>Step 7:</strong> For the equation to have an integer solution, assume $a = b$, so:</p><p>$$(2n+1) \cdot 2 = 31$$</p><p>This fails. Instead, recognize that $(2n+1)$ must divide 31. Since 31 is prime: $2n+1 = 31 \Rightarrow n = 15$</p><p>∴ Answer: C (n = 15)</p>