What sum is obtained by adding \((29+30+31)\) to \((1+2+\cdots+28)\)?
Answer and explanation
Correct answer: 496
The first expression contains the consecutive natural numbers from 1 through 28, and the second expression adds 29, 30, and 31. Together they contain every natural number from 1 through 31 exactly once. Thus the total is \(S_{31}=31\times32/2=31\times16=496\). Therefore option A is correct. It is unnecessary to calculate the two parts separately, although doing so gives \(S_{28}=406\) and \(29+30+31=90\), whose sum is also 496. Options B, C, and D reflect errors such as using the wrong final value, miscounting the added terms, or performing the arithmetic incorrectly.
Frequently asked questions
What is the correct answer to this question?
496
Why is this the correct answer?
The first expression contains the consecutive natural numbers from 1 through 28, and the second expression adds 29, 30, and 31. Together they contain every natural number from 1 through 31 exactly once. Thus the total is \(S_{31}=31\times32/2=31\times16=496\). Therefore option A is correct. It is unnecessary to calculate the two parts separately, although doing so gives \(S_{28}=406\) and \(29+30+31=90\), whose sum is also 496. Options B, C, and D reflect errors such as using the wrong final value, miscounting the added terms, or performing the arithmetic incorrectly.
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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