What is the general term of the sequence (8,27,56,95,\ldots)?
Answer and explanation
Correct answer: \(a_n=5n^2+4n-1\)
The consecutive differences are \(19,29,39\), whose second differences are \(10\). Therefore, the sequence has a quadratic general term of the form \(a_n=5n^2+bn+c\), since its second difference is \(2\times5=10\). Substituting \(n=1\) and \(n=2\) gives \(b=4\) and \(c=-1\), so \(a_n=5n^2+4n-1\) is correct. The close distractor \(a_n=5n^2+2n+1\) gives the first term correctly but gives \(25\), not \(27\), for the second term. Exam tip: verify a proposed general term using at least the first two or three terms.
Frequently asked questions
What is the correct answer to this question?
\(a_n=5n^2+4n-1\)
Why is this the correct answer?
The consecutive differences are \(19,29,39\), whose second differences are \(10\). Therefore, the sequence has a quadratic general term of the form \(a_n=5n^2+bn+c\), since its second difference is \(2\times5=10\). Substituting \(n=1\) and \(n=2\) gives \(b=4\) and \(c=-1\), so \(a_n=5n^2+4n-1\) is correct. The close distractor \(a_n=5n^2+2n+1\) gives the first term correctly but gives \(25\), not \(27\), for the second term. Exam tip: verify a proposed general term using at least the first two or three terms.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Explicit or general rule.
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