What is the sum of \(121+122+123+\cdots+160\)?
Answer and explanation
Correct answer: 5620
The governing idea is to express a consecutive range as the difference of two initial natural-number sums. The required terms are 121 through 160, so their sum is \(S_{160}-S_{120}\). Using \(S_n=\frac{n(n+1)}2\), we get \(S_{160}=\frac{160\cdot161}{2}=12880\) and \(S_{120}=\frac{120\cdot121}{2}=7260\). Therefore the required sum is \(12880-7260=5620\), so option D is correct. Directly counting 40 terms and using the arithmetic-series formula gives the same result: \(40(121+160)/2=20\cdot281=5620\).
Frequently asked questions
What is the correct answer to this question?
5620
Why is this the correct answer?
The governing idea is to express a consecutive range as the difference of two initial natural-number sums. The required terms are 121 through 160, so their sum is \(S_{160}-S_{120}\). Using \(S_n=\frac{n(n+1)}2\), we get \(S_{160}=\frac{160\cdot161}{2}=12880\) and \(S_{120}=\frac{120\cdot121}{2}=7260\). Therefore the required sum is \(12880-7260=5620\), so option D is correct. Directly counting 40 terms and using the arithmetic-series formula gives the same result: \(40(121+160)/2=20\cdot281=5620\).
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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