<p>What is the maximum sum of the series \[20 + 19\frac{1}{3} + 18\frac{2}{3} + \ldots?\]</p>
Step-by-Step Solution
Key Concept: This is an arithmetic sequence with negative common difference (d = -2/3). The series reaches maximum sum when we stop adding terms right before they become negative, since adding negative terms decreases the total.
<p><strong>Step 1:</strong> Identify the sequence. First term a₁ = 20, common difference d = 19⅓ - 20 = -⅔</p><p><strong>Step 2:</strong> Find when terms become negative. The nth term is: aₙ = 20 + (n-1)(-⅔) = 20 - (2n-2)/3 = (60-2n+2)/3 = (62-2n)/3</p><p><strong>Step 3:</strong> Set aₙ ≥ 0: (62-2n)/3 ≥ 0 → 62-2n ≥ 0 → n ≤ 31. The last positive term is a₃₁.</p><p><strong>Step 4:</strong> Verify a₃₁ = (62-62)/3 = 0 and a₃₂ = (62-64)/3 = -⅔ < 0. So maximum sum includes terms up to a₃₁ = 0.</p><p><strong>Step 5:</strong> Calculate S₃₁ = (n/2)(a₁ + aₙ) = (31/2)(20 + 0) = (31/2)(20) = 310</p><p>∴ Answer: <strong>310</strong></p>
Correct Answer: 310