If \(S_n-S_{n-6}=621\), where \(S_k=1+2+\cdots+k\), what is the value of \(n\)?
Answer and explanation
Correct answer: 106
Subtracting the two sums cancels the common terms from 1 through \(n-6\). The remaining six terms are \((n-5),(n-4),(n-3),(n-2),(n-1),n\). Their sum is \(6n-(5+4+3+2+1)=6n-15\). Set this equal to the given value: \(6n-15=621\), so \(6n=636\) and \(n=106\). Hence option C is correct. The expression is not \(6n\), because the five subtracted offsets contribute 15. Options A, B, and D give 609, 615, and 627 respectively, so they fail the required equality.
Frequently asked questions
What is the correct answer to this question?
106
Why is this the correct answer?
Subtracting the two sums cancels the common terms from 1 through \(n-6\). The remaining six terms are \((n-5),(n-4),(n-3),(n-2),(n-1),n\). Their sum is \(6n-(5+4+3+2+1)=6n-15\). Set this equal to the given value: \(6n-15=621\), so \(6n=636\) and \(n=106\). Hence option C is correct. The expression is not \(6n\), because the five subtracted offsets contribute 15. Options A, B, and D give 609, 615, and 627 respectively, so they fail the required equality.
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.
