Find the value of (S_{40}+S_{60}-S_{50}).
Answer and explanation
Correct answer: 1375
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{40}=\frac{40\times41}{2}=820\), \(S_{60}=\frac{60\times61}{2}=1830\), and \(S_{50}=\frac{50\times51}{2}=1275\). Therefore, \(S_{40}+S_{60}-S_{50}=820+1830-1275=1375\). An answer such as 1385 can result from an error while evaluating one of the sums. Exam tip: calculate each \(S_n\) separately before performing the final addition and subtraction.
Frequently asked questions
What is the correct answer to this question?
1375
Why is this the correct answer?
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{40}=\frac{40\times41}{2}=820\), \(S_{60}=\frac{60\times61}{2}=1830\), and \(S_{50}=\frac{50\times51}{2}=1275\). Therefore, \(S_{40}+S_{60}-S_{50}=820+1830-1275=1375\). An answer such as 1385 can result from an error while evaluating one of the sums. Exam tip: calculate each \(S_n\) separately before performing the final addition and subtraction.
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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