If (a_n=n^2-3n+5), what is the smallest term value?
Answer and explanation
Correct answer: 3
For positive integer values of n, \(a_n=n^2-3n+5=(n-1)(n-2)+3\). The product \((n-1)(n-2)\) is 0 at n = 1 and n = 2, and it is positive for other positive integers. Hence, the smallest term value is \(3\). Option \(2\) may seem tempting because the quadratic has its real minimum at \(n=\tfrac32\), but a sequence index must be an integer. Exam tip: while finding extrema of a sequence, use only valid integer indices.
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What is the correct answer to this question?
3
Why is this the correct answer?
For positive integer values of n, \(a_n=n^2-3n+5=(n-1)(n-2)+3\). The product \((n-1)(n-2)\) is 0 at n = 1 and n = 2, and it is positive for other positive integers. Hence, the smallest term value is \(3\). Option \(2\) may seem tempting because the quadratic has its real minimum at \(n=\tfrac32\), but a sequence index must be an integer. Exam tip: while finding extrema of a sequence, use only valid integer indices.
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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