<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 value of \(q - p\) 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 each set as (5-D, 5, 5+D) and (5-d, 5, 5+d), then apply the product ratio condition with D - d = 1 to solve for d.
<p><strong>Step 1:</strong> For three numbers in A.P. with sum 15, let them be (a-r), a, (a+r). Then 3a = 15, so a = 5.</p><p><strong>Step 2:</strong> Set A: (5-D), 5, (5+D) with common difference D. Set B: (5-d), 5, (5+d) with common difference d.</p><p><strong>Step 3:</strong> Calculate products:</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 4:</strong> Use the ratio condition p/q = 7/8:</p><p>$$\frac{5(25-D²)}{5(25-d²)} = \frac{7}{8}$$</p><p>$$\frac{25-D²}{25-d²} = \frac{7}{8}$$</p><p><strong>Step 5:</strong> Cross-multiply: 8(25-D²) = 7(25-d²)</p><p>200 - 8D² = 175 - 7d²</p><p>25 = 8D² - 7d²</p><p><strong>Step 6:</strong> Substitute D = d + 1:</p><p>25 = 8(d+1)² - 7d²</p><p>25 = 8(d² + 2d + 1) - 7d²</p><p>25 = 8d² + 16d + 8 - 7d²</p><p>25 = d² + 16d + 8</p><p>d² + 16d - 17 = 0</p><p><strong>Step 7:</strong> Factor: (d + 17)(d - 1) = 0. Since d > 0, we get d = 1.</p><p><strong>Step 8:</strong> Therefore D = 2. Then:</p><p>q = 5(25 - 1) = 5(24) = 120</p><p>p = 5(25 - 4) = 5(21) = 105</p><p>∴ q - p = 120 - 105 = <strong>15</strong></p>
Correct Answer: A