If (S_a=3003) and (S_b=5050), what will be the value of (b-a)?
Answer and explanation
Correct answer: 23
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{a(a+1)}{2}=3003\), we get \(a=77\), since \(\frac{77\times78}{2}=3003\). Similarly, \(\frac{b(b+1)}{2}=5050\) gives \(b=100\). Therefore, \(b-a=100-77=23\). Although 22 is close, it is not the difference between the correct indices. Exam tip: when a sum is given, look for consecutive integers satisfying \(n(n+1)/2\).
Frequently asked questions
What is the correct answer to this question?
23
Why is this the correct answer?
The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{a(a+1)}{2}=3003\), we get \(a=77\), since \(\frac{77\times78}{2}=3003\). Similarly, \(\frac{b(b+1)}{2}=5050\) gives \(b=100\). Therefore, \(b-a=100-77=23\). Although 22 is close, it is not the difference between the correct indices. Exam tip: when a sum is given, look for consecutive integers satisfying \(n(n+1)/2\).
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.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
