If (S_n=20100), what will be the value of (S_n-S_{n-8})?
Answer and explanation
Correct answer: (1572)
For the sum of the first \\(n\\) natural numbers, \\(S_n=\\frac{n(n+1)}2\\). We are told that \\(S_n=20100\\). Solving \\(\\frac{n(n+1)}2=20100\\) gives \\(n(n+1)=40200\\), and \\(200\\times201=40200\\), so \\(n=200\\). Therefore \\(S_n-S_{n-8}\\) is the sum of the last eight numbers through 200.
Those numbers are 193, 194, 195, 196, 197, 198, 199, and 200. Their sum can be found by pairing the first and last terms: there are 8 terms and each pair totals 393, giving \\(8\\times\\frac{193+200}{2}=8\\times196.5=1572\\). Hence option A is correct.
Frequently asked questions
What is the correct answer to this question?
(1572)
Why is this the correct answer?
For the sum of the first \\(n\\) natural numbers, \\(S_n=\\frac{n(n+1)}2\\). We are told that \\(S_n=20100\\). Solving \\(\\frac{n(n+1)}2=20100\\) gives \\(n(n+1)=40200\\), and \\(200\\times201=40200\\), so \\(n=200\\). Therefore \\(S_n-S_{n-8}\\) is the sum of the last eight numbers through 200.
Those numbers are 193, 194, 195, 196, 197, 198, 199, and 200. Their sum can be found by pairing the first and last terms: there are 8 terms and each pair totals 393, giving \\(8\\times\\frac{193+200}{2}=8\\times196.5=1572\\). Hence option A is correct.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Sum of first n natural numbers.
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