If (S_n=11325), what will be (S_n-S_{n-6})?
Answer and explanation
Correct answer: 885
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=11325\), we get \(n=150\). Therefore, \(S_n-S_{n-6}\) is the sum of the last 6 numbers, \(145+146+147+148+149+150=885\). Option 875 is not the correct sum of these six terms. Exam tip: \(S_n-S_{n-r}\) always represents the sum of the last r terms up to n.
Frequently asked questions
What is the correct answer to this question?
885
Why is this the correct answer?
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=11325\), we get \(n=150\). Therefore, \(S_n-S_{n-6}\) is the sum of the last 6 numbers, \(145+146+147+148+149+150=885\). Option 875 is not the correct sum of these six terms. Exam tip: \(S_n-S_{n-r}\) always represents the sum of the last r terms up to n.
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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