Find the value of (S_{180}-S_{170}+S_{20}).
Answer and explanation
Correct answer: 1965
Here, \(S_n=1+2+\cdots+n=\frac{n(n+1)}{2}\). Thus, \(S_{180}=\frac{180\times181}{2}=16290\), \(S_{170}=\frac{170\times171}{2}=14535\), and \(S_{20}=\frac{20\times21}{2}=210\). Therefore, \(S_{180}-S_{170}+S_{20}=16290-14535+210=1965\). The option 1975 may result from an error of 10 during subtraction or addition. Exam tip: \(S_{180}-S_{170}\) can also be treated as the sum of the terms from 171 to 180.
Frequently asked questions
What is the correct answer to this question?
1965
Why is this the correct answer?
Here, \(S_n=1+2+\cdots+n=\frac{n(n+1)}{2}\). Thus, \(S_{180}=\frac{180\times181}{2}=16290\), \(S_{170}=\frac{170\times171}{2}=14535\), and \(S_{20}=\frac{20\times21}{2}=210\). Therefore, \(S_{180}-S_{170}+S_{20}=16290-14535+210=1965\). The option 1975 may result from an error of 10 during subtraction or addition. Exam tip: \(S_{180}-S_{170}\) can also be treated as the sum of the terms from 171 to 180.
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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