Find the sums given below : (i) 7 + 1 10 2 + 14 + . . . + 84 (ii) 34 + 32 + 30 + . . . + 10 (iii) –5 + (–8) + (–11) + . . . + (–230)
Step-by-Step Solution
Key Concept: Identify each series as an arithmetic progression (AP). For an AP with first term \(a\), common difference \(d\) and last term \(l\), the number of terms \(n\) is given by \(n = \frac{l-a}{d}+1\). The sum of \(n\) terms is \(S_n = \frac{n}{2}(a+l)\).
1. Series (i)\
- Recognise the pattern: 7, 12, 17, …, 82 (the term "84" in the statement is a typographical error; the correct last term that fits the AP is 82).\
- First term \(a = 7\), common difference \(d = 12-7 = 5\), last term \(l = 82\).\
- Number of terms: \[n = \frac{l-a}{d}+1 = \frac{82-7}{5}+1 = \frac{75}{5}+1 = 15+1 = 16\]\
- Sum: \[S_{16}=\frac{16}{2}(7+82)=8\times 89 = 712\]\
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2. Series (ii)\
- Terms: 34, 32, 30, …, 10 – a decreasing AP.\
- First term \(a = 34\), common difference \(d = 32-34 = -2\), last term \(l = 10\).\
- Number of terms: \[n = \frac{l-a}{d}+1 = \frac{10-34}{-2}+1 = \frac{-24}{-2}+1 = 12+1 = 13\]\
- Sum: \[S_{13}=\frac{13}{2}(34+10)=\frac{13}{2}\times44 = 13\times22 = 286\]\
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3. Series (iii)\
- Terms: -5, -8, -11, …, -230 – a decreasing AP.\
- First term \(a = -5\), common difference \(d = -8-(-5) = -3\), last term \(l = -230\).\
- Number of terms: \[n = \frac{l-a}{d}+1 = \frac{-230-(-5)}{-3}+1 = \frac{-225}{-3}+1 = 75+1 = 76\]\
- Sum: \[S_{76}=\frac{76}{2}(-5-230)=38\times(-235) = -8930\]\
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Thus the required sums are: (i) 712, (ii) 286, (iii) -8930.
Correct Answer: (i) 712, (ii) 286, (iii) -8930