What is the value of (S_{48}+S_{96}-S_{32})?
Answer and explanation
Correct answer: 5304
Here, \(S_n\) denotes the sum of the first \(n\) natural numbers, so \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{48}=\frac{48\times49}{2}=1176\), \(S_{96}=\frac{96\times97}{2}=4656\), and \(S_{32}=\frac{32\times33}{2}=528\). Therefore, \(S_{48}+S_{96}-S_{32}=1176+4656-528=5304\). The value 5284 can result from an arithmetic error in addition or subtraction. Exam tip: calculate each \(S_n\) separately before carrying out the final operation.
Frequently asked questions
What is the correct answer to this question?
5304
Why is this the correct answer?
Here, \(S_n\) denotes the sum of the first \(n\) natural numbers, so \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{48}=\frac{48\times49}{2}=1176\), \(S_{96}=\frac{96\times97}{2}=4656\), and \(S_{32}=\frac{32\times33}{2}=528\). Therefore, \(S_{48}+S_{96}-S_{32}=1176+4656-528=5304\). The value 5284 can result from an arithmetic error in addition or subtraction. Exam tip: calculate each \(S_n\) separately before carrying out the final operation.
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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