Find the sum of natural numbers from (121) to (150).
Answer and explanation
Correct answer: 4065
There are 30 terms from 121 to 150. This is an arithmetic progression, so
\(S=\frac{n}{2}(a+l)=\frac{30}{2}(121+150)=15\times271=4065\). Therefore, the correct answer is 4065. An answer such as 4035 may result from an error in counting the terms or using the last term. Exam tip: for a sum from \(a\) to \(l\), always count the terms as \(l-a+1\).
Frequently asked questions
What is the correct answer to this question?
4065
Why is this the correct answer?
There are 30 terms from 121 to 150. This is an arithmetic progression, so
\(S=\frac{n}{2}(a+l)=\frac{30}{2}(121+150)=15\times271=4065\). Therefore, the correct answer is 4065. An answer such as 4035 may result from an error in counting the terms or using the last term. Exam tip: for a sum from \(a\) to \(l\), always count the terms as \(l-a+1\).
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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