<p>If \( x, |x+1|, |x-1| \) are the first three terms of an arithmetic progression (in that order), then the sum of the first 20 terms of this arithmetic progression can be:</p>
Step-by-Step Solution
Key Concept: For an A.P., the common difference must be constant, so |x+1| - x = |x-1| - |x+1|. The absolute value expressions require careful case analysis based on the sign of x and x±1.
<p><strong>Step 1: Set up the A.P. condition</strong></p><p>For A.P.: |x+1| - x = |x-1| - |x+1|</p><p>This gives: 2|x+1| = x + |x-1|</p><p><strong>Step 2: Case Analysis</strong></p><p><u>Case 1: x ≥ 1</u></p><p>Then |x+1| = x+1 and |x-1| = x-1</p><p>2(x+1) = x + (x-1) → 2x + 2 = 2x - 1 → Contradiction</p><p><u>Case 2: -1 ≤ x < 1</u></p><p>Then |x+1| = x+1 and |x-1| = 1-x</p><p>2(x+1) = x + (1-x) → 2x + 2 = 1 → x = -1/2</p><p><strong>Step 3: Verify and find the A.P.</strong></p><p>With x = -1/2: First three terms are -1/2, |1/2|, |−3/2| = -1/2, 1/2, 3/2</p><p>Common difference d = 1</p><p><strong>Step 4: Sum of first 20 terms</strong></p><p>S₂₀ = (n/2)[2a + (n-1)d] = (20/2)[2(-1/2) + 19(1)]</p><p>S₂₀ = 10[-1 + 19] = 10(18) = 180</p><p>∴ Answer: A</p>
Correct Answer: A