How much greater is the sum of the first (90) natural numbers than the sum of the first (45) natural numbers?
Answer and explanation
Correct answer: 3060
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{90}=\frac{90\times91}{2}=4095\) and \(S_{45}=\frac{45\times46}{2}=1035\). Therefore, the required difference is \(4095-1035=3060\). The value 4095 is only the sum of the first 90 natural numbers, not the difference. Exam tip: write \(S_{90}-S_{45}\) before calculating in such questions.
Frequently asked questions
What is the correct answer to this question?
3060
Why is this the correct answer?
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{90}=\frac{90\times91}{2}=4095\) and \(S_{45}=\frac{45\times46}{2}=1035\). Therefore, the required difference is \(4095-1035=3060\). The value 4095 is only the sum of the first 90 natural numbers, not the difference. Exam tip: write \(S_{90}-S_{45}\) before calculating 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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