\( -1, \quad 2, \quad 7, \quad 14, \quad 23, \quad \ldots \) a) Work out the \( n^{\text {th }} \) term rule for this sequence. b) At what position does the term with value 62 appear in this sequence?
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To find the \( n^{\text {th }} \) term of the sequence, let's first look at the pattern. The difference between consecutive terms are: \[ 2 - (-1) = 3, \quad 7 - 2 = 5, \quad 14 - 7 = 7, \quad 23 - 14 = 9 \] The differences (3, 5, 7, 9) form an arithmetic sequence with a common difference of 2. Thus, the \( n^{\text{th}} \) term can be expressed as: \[ a_n = a_1 + \text{sum of the first (n-1) terms of the difference} \] The first term \( a_1 = -1 \), the sum of the first \( n - 1 \) terms of the differences is given by: \[ \text{difference sum} = 3 + 5 + 7 + \ldots = (1 + 2 + \ldots + (2n - 3)) + 3(n - 1) = (n-1)^2 + 3(n-1) \] Thus the \( n^{\text{th}} \) term can be simplified to: \[ a_n = -1 + (n-1)^2 + 3(n-1) = n^2 + n - 1 \] So the \( n^{\text{th}} \) term rule for the sequence is: \[ a_n = n^2 + n - 1 \] For the second part, we need to find \( n \) such that \( a_n = 62 \): \[ n^2 + n - 1 = 62 \] \[ n^2 + n - 63 = 0 \] Now we can factor this quadratic equation: \[ (n - 7)(n + 9) = 0 \] Thus, \( n = 7 \) (since \( n \) must be positive) shows that the term with a value of 62 appears at position 7 in the sequence. So, the answers are: a) \( a_n = n^2 + n - 1 \) b) The term with value 62 appears at position 7.