What is the sum of natural numbers from (44) to (97)?
Answer and explanation
Correct answer: (3807)
The required integers are 44, 45, ..., 97. For a range beginning at 44, it is convenient to subtract the sum of integers through 43 from the sum through 97. The formula for the first n natural numbers is \\(S_n=\frac{n(n+1)}{2}\\). Therefore, \\(S_{97}=\frac{97\cdot98}{2}=4753\\), and \\(S_{43}=\frac{43\cdot44}{2}=946\\). Their difference is \\(4753-946=3807\\).
Hence option D is correct. Subtracting \\(S_{44}\\) would remove 44 itself and give the wrong range, so the upper limit before the starting number must be 43. A quick check also helps: there are 54 terms, and their average is \\(\frac{44+97}{2}=70.5\\); multiplying gives \\(54\times70.5=3807\\).
Frequently asked questions
What is the correct answer to this question?
(3807)
Why is this the correct answer?
The required integers are 44, 45, ..., 97. For a range beginning at 44, it is convenient to subtract the sum of integers through 43 from the sum through 97. The formula for the first n natural numbers is \\(S_n=\frac{n(n+1)}{2}\\). Therefore, \\(S_{97}=\frac{97\cdot98}{2}=4753\\), and \\(S_{43}=\frac{43\cdot44}{2}=946\\). Their difference is \\(4753-946=3807\\).
Hence option D is correct. Subtracting \\(S_{44}\\) would remove 44 itself and give the wrong range, so the upper limit before the starting number must be 43. A quick check also helps: there are 54 terms, and their average is \\(\frac{44+97}{2}=70.5\\); multiplying gives \\(54\times70.5=3807\\).
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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