<p>The terms \(a_1, a_2, a_3\) form an arithmetic sequence whose sum is 18. The terms \(a_1 + 1, a_2, a_3 + 2\), in that order, form a geometric sequence. Then the sum of all possible common difference of the A.P. is ___.</p>
Step-by-Step Solution
Key Concept: Set up equations from AP sum condition and GP ratio condition (consecutive terms have equal ratios), then solve the resulting quadratic in the common difference to find all possible values.
<p><strong>Step 1:</strong> Let the AP be: a₁ = a, a₂ = a+d, a₃ = a+2d, where d is the common difference.</p><p><strong>Step 2:</strong> From sum condition: a + (a+d) + (a+2d) = 18 → 3a + 3d = 18 → a + d = 6 → a = 6 - d</p><p><strong>Step 3:</strong> The sequence a₁+1, a₂, a₃+2 becomes: (a+1), (a+d), (a+2d+2) which form a GP.</p><p><strong>Step 4:</strong> For GP, the ratio condition gives: (a+d)/(a+1) = (a+2d+2)/(a+d)</p><p><strong>Step 5:</strong> Cross multiply: (a+d)² = (a+1)(a+2d+2)</p><p><strong>Step 6:</strong> Substitute a = 6-d: (6-d+d)² = (6-d+1)(6-d+2d+2) → 36 = (7-d)(8+d)</p><p><strong>Step 7:</strong> Expand: 36 = 56 + 7d - 8d - d² → d² + d - 20 = 0</p><p><strong>Step 8:</strong> Factor: (d+5)(d-4) = 0 → d = -5 or d = 4</p><p><strong>Step 9:</strong> Sum of all possible common differences = -5 + 4 = <strong>-1</strong></p><p><strong>Verification:</strong> For d=4: a=2, AP is 2,6,10 (sum=18✓), GP: 3,6,12 (ratios 2,2✓). For d=-5: a=11, AP is 11,6,1 (sum=18✓), GP: 12,6,3 (ratios 0.5,0.5✓)</p><p>∴ Answer: <strong>-1</strong></p>
Correct Answer: 6