QUESTION 2 [7] The sequence \( 4 ; 9 ; x ; 37 ; \ldots \) is a quadratic sequence. 2.1 Calculate \( x \). (3) 2.2 Hence, or otherwise, determine the \( n^{\text {th }} \) term of the sequence. (4)
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To find \( x \) in the sequence \( 4, 9, x, 37, \ldots \), we first identify the differences between the consecutive terms. The first difference is \( 9 - 4 = 5 \) and the second difference would be the difference between \( x \) and \( 9 \), followed by the difference between \( 37 \) and \( x \). Since it’s a quadratic sequence, the second difference will be constant. Let’s denote the first differences as \( d_1 = 5 \), \( d_2 = x - 9 \), and \( d_3 = 37 - x \). To ensure a constant second difference, we would have: \( d_2 - d_1 = d_3 - d_2 \). This leads us to the equation: \[ (x - 9) - 5 = (37 - x) - (x - 9). \] Simplifying: \[ x - 14 = 46 - 2x. \] Bringing all terms involving \( x \) to one side gives: \[ 3x = 60 \quad \Rightarrow \quad x = 20. \] Thus, \( x = 20 \). Now for the \( n^{\text{th}} \) term, the sequence can be generated using the first term and the pattern in the differences: - First differences are: \( 5, 11, 17, \ldots \) (which is a linear sequence with a constant difference of 6). - Second differences are constant at 6. With a general form of the quadratic equation \( a_n = an^2 + bn + c \), we can set up equations based on known values: For \( n = 1: a(1^2) + b(1) + c = 4 \) leads to \( a + b + c = 4 \). For \( n = 2: a(2^2) + b(2) + c = 9 \) simplifies to \( 4a + 2b + c = 9 \). For \( n = 4: a(4^2) + b(4) + c = 37 \) simplifies to \( 16a + 4b + c = 37 \). From these equations, we can solve for \( a, b, c \): 1. \( a + b + c = 4 \) 2. \( 4a + 2b + c = 9 \) 3. \( 16a + 4b + c = 37 \) By solving these simultaneously, we can find \( a = 1 \), \( b = 2 \), and \( c = 1 \). Therefore, the \( n^{\text{th}} \) term is \( a_n = n^2 + 2n + 1 = (n + 1)^2 \).
