Which is the (n)th term of the sequence (0,5,16,33,56,\ldots)?
Answer and explanation
Correct answer: \(3n^2-4n+1\)
The first differences are \(5,11,17,23\), and the second differences are all \(6\). Hence the sequence has a quadratic nth term, with coefficient of \(n^2\) equal to \(6/2=3\). For \(a_n=3n^2-4n+1\), we get \(a_1=0\), \(a_2=5\), \(a_3=16\), and \(a_4=33\), so option A is correct. Option C gives \(a_1=0\), but it gives \(a_2=6\), not \(5\). Exam tip: when second differences are constant, start with \(an^2+bn+c\).
Frequently asked questions
What is the correct answer to this question?
\(3n^2-4n+1\)
Why is this the correct answer?
The first differences are \(5,11,17,23\), and the second differences are all \(6\). Hence the sequence has a quadratic nth term, with coefficient of \(n^2\) equal to \(6/2=3\). For \(a_n=3n^2-4n+1\), we get \(a_1=0\), \(a_2=5\), \(a_3=16\), and \(a_4=33\), so option A is correct. Option C gives \(a_1=0\), but it gives \(a_2=6\), not \(5\). Exam tip: when second differences are constant, start with \(an^2+bn+c\).
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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