<p><strong>For Problems 16–18:</strong> There are two sets \(A\) and \(B\) each of which consists of three numbers in A.P. whose sum is 15 and where \(D\) and \(d\) are the common differences such that \(D - d = 1\). If \(\frac{p}{q} = \frac{7}{8}\), where \(p\) and \(q\) are the product of the numbers, respectively, and \(d > 0\) in the two sets.</p><p>The sum of the product of the numbers in set \(A\) taken two at a time is</p>
Step-by-Step Solution
Key Concept: For three numbers in A.P. with sum 15, the middle term is always 5. Use the constraint D - d = 1 and the product ratio p/q = 7/8 to find the specific common differences, then calculate the sum of products taken two at a time using the identity (a-D)(5)(a+D) and similar expressions.
<p><strong>Step 1:</strong> Let set A have terms (5-D), 5, (5+D) and set B have terms (5-d), 5, (5+d), where D and d are common differences with D - d = 1.</p><p><strong>Step 2:</strong> Both sets have sum = (5-D) + 5 + (5+D) = 15 ✓ and (5-d) + 5 + (5+d) = 15 ✓</p><p><strong>Step 3:</strong> Product of set A: p = (5-D)·5·(5+D) = 5(25-D²)</p><p>Product of set B: q = (5-d)·5·(5+d) = 5(25-d²)</p><p><strong>Step 4:</strong> From p/q = 7/8: $\frac{5(25-D²)}{5(25-d²)} = \frac{7}{8}$ → $\frac{25-D²}{25-d²} = \frac{7}{8}$</p><p><strong>Step 5:</strong> Cross-multiply: 8(25-D²) = 7(25-d²) → 200 - 8D² = 175 - 7d²</p><p>→ 25 = 8D² - 7d²</p><p><strong>Step 6:</strong> Substitute D = d + 1: 25 = 8(d+1)² - 7d² = 8d² + 16d + 8 - 7d²</p><p>→ d² + 16d + 8 = 25 → d² + 16d - 17 = 0</p><p><strong>Step 7:</strong> (d + 17)(d - 1) = 0 → d = 1 (since d > 0), so D = 2</p><p><strong>Step 8:</strong> Set A: 3, 5, 7. Sum of products taken two at a time = (3)(5) + (5)(7) + (7)(3) = 15 + 35 + 21 = <strong>71</strong></p><p>∴ Answer: C</p>
Correct Answer: C