Question:

Step 1: Set up the arithmetic progression with odd number of terms. Let the AP have $n = 2m+1$ terms (where $n$ is odd), with first term $a$ and common difference $d > 0$ (positive integer). Step 2: Identify the odd-positioned terms and find their sum. The odd-positioned terms are the 1st, 3rd, 5th, 7th, ... terms. These form a new AP with: - First term: $a$ - Common difference: $2d$ - Number of terms: $m+1$ The sum of odd-positioned terms is: $$S_1 = (m+1)a + 2d \cdot \frac{m(m+1)}{2} = (m+1)a + md(m+1) = (m+1)(a+md)$$ Note that $a+md$ is the middle term of the original AP. Step 3: Find the total sum of the AP. The sum of all $n = 2m+1$ terms is: $$S = \frac{n}{2}(2a + (n-1)d) = \frac{2m+1}{2}(2a + 2md) = (2m+1)(a+md)$$ Step 4: Calculate the sum of remaining (even-positioned) terms. The sum of the remaining terms (even-positioned terms) is: $$S_2 = S - S_1 = (2m+1)(a+md) - (m+1)(a+md)$$ $$S_2 = (a+md)[(2m+1) - (m+1)] = (a+md) \cdot m = m(a+md)$$ Step 5: Set up the ratio equation using the given condition. We are given that: $$\frac{S_1}{S_2} = \frac{13}{12}$$ Substituting our expressions: $$\frac{(m+1)(a+md)}{m(a+md)} = \frac{13}{12}$$ Since $a+md \neq 0$ (the AP is non-constant), we can cancel it: $$\frac{m+1}{m} = \frac{13}{12}$$ Step 6: Solve for $m$ and find the number of terms. Cross-multiplying: $$12(m+1) = 13m$$ $$12m + 12 = 13m$$ $$m = 12$$ Therefore, the number of terms is: $$n = 2m + 1 = 2(12) + 1 = 25$$ The answer is **Option 3: 25**