Which is the (n)th term of the sequence (4,7,14,25,40,\ldots)?
Answer and explanation
Correct answer: \(2n^2-3n+5\)
The first differences are \(3,7,11,15\), and the second differences are all \(4\). Hence, the nth term has the quadratic form \(an^2+bn+c\). Since \(2a=4\), we get \(a=2\). Using the first two terms gives \(b=-3\) and \(c=5\), so the nth term is \(2n^2-3n+5\). Option B has the same leading coefficient, but although it gives 4 for \(n=1\), it gives 8 for \(n=2\), not 7. Exam tip: when second differences are constant, assume a quadratic expression and determine its coefficients from initial terms.
Frequently asked questions
What is the correct answer to this question?
\(2n^2-3n+5\)
Why is this the correct answer?
The first differences are \(3,7,11,15\), and the second differences are all \(4\). Hence, the nth term has the quadratic form \(an^2+bn+c\). Since \(2a=4\), we get \(a=2\). Using the first two terms gives \(b=-3\) and \(c=5\), so the nth term is \(2n^2-3n+5\). Option B has the same leading coefficient, but although it gives 4 for \(n=1\), it gives 8 for \(n=2\), not 7. Exam tip: when second differences are constant, assume a quadratic expression and determine its coefficients from 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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