What is (40%) of the sum of the first (125) natural numbers?
Answer and explanation
Correct answer: (3150)
The sum of the first n natural numbers is \\(S_n=\frac{n(n+1)}{2}\\). For n=125, \\(S_{125}=\frac{125\times126}{2}=125\times63=7875\\). The question asks for 40 percent of this sum. Since \\(40\%=\frac{40}{100}=\frac25\\), multiply \\(7875\\) by \\(\frac25\\): \\(7875\times\frac25=1575\times2=3150\\).
Therefore option A is correct. The percentage must be applied to the entire sum, not just to 125, and 40 percent is not the same as multiplying by 40. Converting the percentage to a fraction makes the calculation clear and avoids a decimal error. The result is an integer and agrees with the listed first option.
Frequently asked questions
What is the correct answer to this question?
(3150)
Why is this the correct answer?
The sum of the first n natural numbers is \\(S_n=\frac{n(n+1)}{2}\\). For n=125, \\(S_{125}=\frac{125\times126}{2}=125\times63=7875\\). The question asks for 40 percent of this sum. Since \\(40\%=\frac{40}{100}=\frac25\\), multiply \\(7875\\) by \\(\frac25\\): \\(7875\times\frac25=1575\times2=3150\\).
Therefore option A is correct. The percentage must be applied to the entire sum, not just to 125, and 40 percent is not the same as multiplying by 40. Converting the percentage to a fraction makes the calculation clear and avoids a decimal error. The result is an integer and agrees with the listed first option.
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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