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What is obtained by subtracting the sum of the first (23) natural numbers from the sum of the first (41) natural numbers?

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Answer and explanation

Correct answer: 585

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{41}=\frac{41\times42}{2}=861\) and \(S_{23}=\frac{23\times24}{2}=276\). Therefore, \(S_{41}-S_{23}=861-276=585\). This is also the sum of the natural numbers from 24 to 41. Option 575 is only a nearby value, not the correct difference. Exam tip: You may also directly find the sum from 24 to 41 in such questions.

Related tags

MathematicsSequences And ProgressionsNatural NumbersSum Of Natural NumbersArithmetic Series

Frequently asked questions

What is the correct answer to this question?

585

Why is this the correct answer?

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{41}=\frac{41\times42}{2}=861\) and \(S_{23}=\frac{23\times24}{2}=276\). Therefore, \(S_{41}-S_{23}=861-276=585\). This is also the sum of the natural numbers from 24 to 41. Option 575 is only a nearby value, not the correct difference. Exam tip: You may also directly find the sum from 24 to 41 in such questions.

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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