What is the general term of the sequence (2,14,51,140,\ldots)?
Answer and explanation
Correct answer: \(a_n=\frac{9n^3-29n^2+48n-24}{2}\)
The first differences are \(12,37,89\), and the second differences are \(25,52\). The third difference is constant at \(27\), so an appropriate rule is a cubic polynomial. Substituting \(n=1,2,3,4\) in \(a_n=\frac{9n^3-29n^2+48n-24}{2}\) gives \(2,14,51,140\), respectively. Option B gives \(42\) when \(n=3\), not \(51\). Exam tip: verify a cubic-type rule by substituting at least the first three or four values of \(n\).
Frequently asked questions
What is the correct answer to this question?
\(a_n=\frac{9n^3-29n^2+48n-24}{2}\)
Why is this the correct answer?
The first differences are \(12,37,89\), and the second differences are \(25,52\). The third difference is constant at \(27\), so an appropriate rule is a cubic polynomial. Substituting \(n=1,2,3,4\) in \(a_n=\frac{9n^3-29n^2+48n-24}{2}\) gives \(2,14,51,140\), respectively. Option B gives \(42\) when \(n=3\), not \(51\). Exam tip: verify a cubic-type rule by substituting at least the first three or four values of \(n\).
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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