If (S_a=820) and (S_b=1035), what is (b-a)?
Answer and explanation
Correct answer: 5
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{a(a+1)}{2}=820\), we get \(a=40\), since \(\frac{40\times41}{2}=820\). Similarly, \(\frac{b(b+1)}{2}=1035\) gives \(b=45\), since \(\frac{45\times46}{2}=1035\). Therefore, \(b-a=45-40=5\). Option 4 may seem close, but it does not equal the actual difference between the indices. Exam tip: equate the given sum to \(\frac{n(n+1)}{2}\) to find the index n.
Frequently asked questions
What is the correct answer to this question?
5
Why is this the correct answer?
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{a(a+1)}{2}=820\), we get \(a=40\), since \(\frac{40\times41}{2}=820\). Similarly, \(\frac{b(b+1)}{2}=1035\) gives \(b=45\), since \(\frac{45\times46}{2}=1035\). Therefore, \(b-a=45-40=5\). Option 4 may seem close, but it does not equal the actual difference between the indices. Exam tip: equate the given sum to \(\frac{n(n+1)}{2}\) to find the index n.
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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