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Subtract the sum of the first (20) natural numbers from the sum of the first (60) natural numbers. What is the result?

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Answer and explanation

Correct answer: 1620

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{60}=\frac{60\times61}{2}=1830\) and \(S_{20}=\frac{20\times21}{2}=210\). Therefore, the required difference is \(1830-210=1620\). It is also the sum of the natural numbers from 21 to 60. Exam tip: when subtracting two partial sums, check the range of terms left over.

Related tags

Natural NumbersSum FormulaSequences And ProgressionsPartial SumsArithmetic Series

Frequently asked questions

What is the correct answer to this question?

1620

Why is this the correct answer?

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{60}=\frac{60\times61}{2}=1830\) and \(S_{20}=\frac{20\times21}{2}=210\). Therefore, the required difference is \(1830-210=1620\). It is also the sum of the natural numbers from 21 to 60. Exam tip: when subtracting two partial sums, check the range of terms left over.

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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