<p>The sum of four whole numbers in AP is 24 and their product is 945; find the numbers.</p>
Step-by-Step Solution
Key Concept: For four numbers in AP with even count, use symmetric form a-3d, a-d, a+d, a+3d so their sum directly gives the middle term, and their product becomes a quadratic in d².
<p><strong>Step 1:</strong> Let the four numbers in AP be (a-3d), (a-d), (a+d), (a+3d) where a, d are whole numbers.</p><p><strong>Step 2:</strong> Sum condition: (a-3d) + (a-d) + (a+d) + (a+3d) = 24</p><p>⟹ 4a = 24 ⟹ a = 6</p><p><strong>Step 3:</strong> Product condition: (a-3d)(a-d)(a+d)(a+3d) = 945</p><p>Rearrange: [(a-3d)(a+3d)][(a-d)(a+d)] = 945</p><p>⟹ (a² - 9d²)(a² - d²) = 945</p><p><strong>Step 4:</strong> Substitute a = 6: (36 - 9d²)(36 - d²) = 945</p><p>Let u = d²: (36 - 9u)(36 - u) = 945</p><p>⟹ 1296 - 36u - 324u + 9u² = 945</p><p>⟹ 9u² - 360u + 351 = 0</p><p>⟹ u² - 40u + 39 = 0</p><p>⟹ (u - 1)(u - 39) = 0</p><p><strong>Step 5:</strong> So u = 1 or u = 39, giving d = ±1 or d = ±√39</p><p>Since d must make whole numbers: d = 1 (checking: √39 is irrational)</p><p><strong>Step 6:</strong> The four numbers are: (6-3), (6-1), (6+1), (6+3) = 3, 5, 7, 9</p><p><strong>Verification:</strong> Sum = 3+5+7+9 = 24 ✓ | Product = 3×5×7×9 = 945 ✓</p><p>∴ Answer: <strong>3, 5, 7, 9</strong></p>
Correct Answer: 3