Which is the (n)th term of the sequence (2,11,28,53,86,\ldots)?
Answer and explanation
Correct answer: \(4n^2-3n+1\)
The first differences are \(9,17,25,33\), and their second differences are all \(8\). Hence the sequence has a quadratic nth term, with coefficient of \(n^2\) equal to \(8/2=4\). Substituting \(n=1,2,3\) in \(4n^2-3n+1\) gives \(2,11,28\), respectively, so it is correct. Although \(4n^2-2n\) gives the first term as \(2\), it gives \(12\) as the second term. Exam tip: when second differences are constant, test a form \(an^2+bn+c\) using the initial terms.
Frequently asked questions
What is the correct answer to this question?
\(4n^2-3n+1\)
Why is this the correct answer?
The first differences are \(9,17,25,33\), and their second differences are all \(8\). Hence the sequence has a quadratic nth term, with coefficient of \(n^2\) equal to \(8/2=4\). Substituting \(n=1,2,3\) in \(4n^2-3n+1\) gives \(2,11,28\), respectively, so it is correct. Although \(4n^2-2n\) gives the first term as \(2\), it gives \(12\) as the second term. Exam tip: when second differences are constant, test a form \(an^2+bn+c\) using the initial terms.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: nth term.
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