<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 \(B\) 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 this to express elements as (5-D, 5, 5+D) and (5-d, 5, 5+d), then apply the product ratio condition to find d, and finally compute the sum of pairwise products.
<p><strong>Step 1: Express sets A and B in A.P. form</strong></p><p>Let set A consist of: (5-D, 5, 5+D) and set B consist of: (5-d, 5, 5+d)</p><p>Both have sum = (5-D) + 5 + (5+D) = 15 ✓</p><p><strong>Step 2: Find products p and q</strong></p><p>p = (5-D)(5)(5+D) = 5(25-D²)</p><p>q = (5-d)(5)(5+d) = 5(25-d²)</p><p><strong>Step 3: Apply the ratio condition</strong></p><p>Given: p/q = 7/8</p><p>Therefore: 5(25-D²)/5(25-d²) = 7/8</p><p>(25-D²)/(25-d²) = 7/8</p><p>8(25-D²) = 7(25-d²)</p><p>200 - 8D² = 175 - 7d²</p><p><strong>Step 4: Use D - d = 1</strong></p><p>Substitute D = d + 1:</p><p>200 - 8(d+1)² = 175 - 7d²</p><p>200 - 8(d² + 2d + 1) = 175 - 7d²</p><p>200 - 8d² - 16d - 8 = 175 - 7d²</p><p>192 - 8d² - 16d = 175 - 7d²</p><p>17 = d² + 16d</p><p>d² + 16d - 17 = 0</p><p>(d + 17)(d - 1) = 0</p><p>Since d > 0: <strong>d = 1</strong></p><p><strong>Step 5: Find sum of pairwise products in set B</strong></p><p>Set B: (4, 5, 6)</p><p>Sum of products taken two at a time = (4)(5) + (5)(6) + (4)(6)</p><p>= 20 + 30 + 24</p><p>= <strong>74</strong></p><p>∴ Answer: D</p>
Correct Answer: D