What is the (n)th term of the sequence (5,18,43,80,129,\ldots)?
Answer and explanation
Correct answer: \(6n^2-5n+4\)
The first differences are \(13,25,37,49\). Their second differences are all \(12\), so the nth term is quadratic and the coefficient of \(n^2\) is \(12/2=6\). Substituting \(n=1,2,3\) in option B gives \(5,18,43\), respectively; hence \(T_n=6n^2-5n+4\). Although option C gives \(5\) when \(n=1\), it gives \(19\) when \(n=2\), so it is incorrect. Exam tip: when second differences are constant, assume the term has the form \(an^2+bn+c\).
Frequently asked questions
What is the correct answer to this question?
\(6n^2-5n+4\)
Why is this the correct answer?
The first differences are \(13,25,37,49\). Their second differences are all \(12\), so the nth term is quadratic and the coefficient of \(n^2\) is \(12/2=6\). Substituting \(n=1,2,3\) in option B gives \(5,18,43\), respectively; hence \(T_n=6n^2-5n+4\). Although option C gives \(5\) when \(n=1\), it gives \(19\) when \(n=2\), so it is incorrect. Exam tip: when second differences are constant, assume the term has the form \(an^2+bn+c\).
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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