<p>Which term of the sequence \( 25, 22\dfrac{3}{4}, 20\dfrac{1}{2}, 18\dfrac{1}{4}, \ldots \) is numerically smallest?</p>
Step-by-Step Solution
Key Concept: Convert to improper fractions to identify the common difference (-9/4), then use the AP formula aₙ = a₁ + (n-1)d to find when terms transition from positive to negative, which determines the smallest term.
<p><strong>Step 1:</strong> Convert sequence to improper fractions: 25, 91/4, 41/2, 73/4, ...</p><p><strong>Step 2:</strong> Find common difference: d = 91/4 - 25 = 91/4 - 100/4 = -9/4</p><p><strong>Step 3:</strong> General term formula: aₙ = 25 + (n-1)(-9/4) = 25 - 9(n-1)/4 = (100 - 9n + 9)/4 = (109 - 9n)/4</p><p><strong>Step 4:</strong> Terms are positive when 109 - 9n > 0, so n < 109/9 ≈ 12.11</p><p><strong>Step 5:</strong> Check n = 12: a₁₂ = (109 - 108)/4 = 1/4 (positive)</p><p><strong>Step 6:</strong> Check n = 13: a₁₃ = (109 - 117)/4 = -8/4 = -2 (negative)</p><p><strong>Step 7:</strong> Compare magnitudes: |1/4| = 0.25 and |-2| = 2, so 1/4 is numerically smaller</p><p>∴ Answer: The 12th term is numerically smallest with value 1/4</p>
Correct Answer: A