What is (4) times the sum of the first (21) natural numbers?
Answer and explanation
Correct answer: 924
The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(S_{21}=\frac{21\times22}{2}=231\). Therefore, the required value is \(4\times231=924\). A choice such as \(904\) may result from an error while finding the sum or multiplying it. Exam tip: calculate \(n(n+1)/2\) first, then multiply by the given factor.
Frequently asked questions
What is the correct answer to this question?
924
Why is this the correct answer?
The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(S_{21}=\frac{21\times22}{2}=231\). Therefore, the required value is \(4\times231=924\). A choice such as \(904\) may result from an error while finding the sum or multiplying it. Exam tip: calculate \(n(n+1)/2\) first, then multiply by the given factor.
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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