Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If the sum of the series \(20 + 19\frac{3}{5} + 19\frac{1}{5} + 18\frac{4}{5} + \ldots\) upto nth term is 488 and the nth term is negative, then find the nth term and n.</p>
<p>(a) nth term is \(-4\frac{2}{5}\)</p>
<p>(b) nth term is \(-4\frac{2}{5}\)</p>
<p>(c) \(n = 60\)</p>
<p>(d) \(n = 41\)</p>
Step-by-Step Solution
Key Concept: Recognize the series as an arithmetic progression with fractional common difference. Use sum formula and the condition that nth term is negative to determine which value of n is valid.
<p><strong>Given:</strong> Series is $20 + 19\frac{3}{5} + 19\frac{1}{5} + 18\frac{4}{5} + \ldots$ with sum = 488 and nth term is negative.</p><p><strong>Step 1:</strong> Convert to improper fractions: $20, \frac{98}{5}, \frac{96}{5}, \frac{94}{5}, \ldots$</p><p><strong>Step 2:</strong> This is an AP with first term $a = 20$ and common difference $d = \frac{98}{5} - 20 = \frac{-2}{5}$.</p><p><strong>Step 3:</strong> Sum formula: $S_n = \frac{n}{2}(2a + (n-1)d) = 488$</p><p>$\frac{n}{2}(40 + (n-1)\frac{-2}{5}) = 488$</p><p>$n(40 - \frac{2(n-1)}{5}) = 976$</p><p>$200n - 2n(n-1) = 4880$</p><p>$n^2 - 101n + 2440 = 0$</p><p><strong>Step 4:</strong> Solving: $n = 40$ or $n = 61$</p><p><strong>Step 5:</strong> For $n = 40$: $T_{40} = 20 + 39 \times \frac{-2}{5} = 20 - \frac{78}{5} = \frac{2}{5}$ (positive)</p><p>For $n = 41$: $T_{41} = 20 + 40 \times \frac{-2}{5} = 20 - 16 = 4$ (positive)</p><p>For $n = 60$: $T_{60} = 20 + 59 \times \frac{-2}{5} = 20 - \frac{118}{5} = \frac{-18}{5}$ (negative)</p><p>For $n = 61$: $T_{61} = 20 + 60 \times \frac{-2}{5} = 20 - 24 = -4\frac{2}{5}$ (negative) ✓</p><p>∴ nth term is $-4\frac{2}{5}$ and $n = 61$</p>
Correct Answer: A