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If (S_n-S_{n-2}=199), what will be the value of (S_n)?

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Answer and explanation

Correct answer: 5050

Let \(S_n\) be the sum of the first \(n\) natural numbers. In \(S_n-S_{n-2}\), only the last two terms, \((n-1)\) and \(n\), remain. Hence, \(2n-1=199\), giving \(n=100\). Therefore, \(S_{100}=\frac{100\times101}{2}=5050\). Option 4950 is the sum of the first 99 natural numbers, so it is not correct. Exam tip: interpret \(S_n-S_{n-2}\) directly as the sum of the last two terms.

Tags

sequences and progressionssum of natural numberspartial sumsalgebraic reasoningclass 9 mathematics

Frequently asked questions

What is the correct answer to this question?

5050

Why is this the correct answer?

Let \(S_n\) be the sum of the first \(n\) natural numbers. In \(S_n-S_{n-2}\), only the last two terms, \((n-1)\) and \(n\), remain. Hence, \(2n-1=199\), giving \(n=100\). Therefore, \(S_{100}=\frac{100\times101}{2}=5050\). Option 4950 is the sum of the first 99 natural numbers, so it is not correct. Exam tip: interpret \(S_n-S_{n-2}\) directly as the sum of the last two terms.

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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