Pregunta

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)

Ask by Schneider Cummings. in South Africa
Mar 09,2025

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\( x = 20 \) and the \( n^\text{th} \) term of the sequence is \( a_n = 3n^2 - 4n + 5 \).

Solución

**Step 1. Calculate \( x \) using the constant second difference property** Since the sequence is quadratic, the second differences are constant. Let the given terms be: \[ a_1 = 4,\quad a_2 = 9,\quad a_3 = x,\quad a_4 = 37. \] First, compute the first differences: \[ d_1 = a_2 - a_1 = 9 - 4 = 5, \] \[ d_2 = a_3 - a_2 = x - 9, \] \[ d_3 = a_4 - a_3 = 37 - x. \] For a quadratic sequence, the second differences are equal: \[ d_2 - d_1 = d_3 - d_2. \] Substitute the known values: \[ (x - 9) - 5 = (37 - x) - (x - 9). \] Simplify both sides: \[ x - 14 = 37 - x - x + 9 = 46 - 2x. \] Solve for \( x \): \[ x - 14 = 46 - 2x, \] \[ x + 2x = 46 + 14, \] \[ 3x = 60, \] \[ x = \frac{60}{3} = 20. \] **Step 2. Determine the \( n^\text{th} \) term of the sequence** Assume the \( n^\text{th} \) term has the form: \[ a_n = An^2 + Bn + C. \] We have: \[ a_1 = A(1)^2 + B(1) + C = A + B + C = 4, \] \[ a_2 = A(2)^2 + B(2) + C = 4A + 2B + C = 9, \] \[ a_3 = A(3)^2 + B(3) + C = 9A + 3B + C = 20. \] Now, set up the equations: 1. \( A + B + C = 4 \), 2. \( 4A + 2B + C = 9 \), 3. \( 9A + 3B + C = 20 \). Subtract the first equation from the second: \[ (4A + 2B + C) - (A + B + C) = 9 - 4, \] \[ 3A + B = 5. \quad \text{(Equation 1)} \] Subtract the second equation from the third: \[ (9A + 3B + C) - (4A + 2B + C) = 20 - 9, \] \[ 5A + B = 11. \quad \text{(Equation 2)} \] Subtract Equation 1 from Equation 2: \[ (5A + B) - (3A + B) = 11 - 5, \] \[ 2A = 6, \] \[ A = 3. \] Substitute \( A = 3 \) into Equation 1: \[ 3(3) + B = 5, \] \[ 9 + B = 5, \] \[ B = 5 - 9 = -4. \] Now, substitute \( A = 3 \) and \( B = -4 \) into the first equation: \[ 3 - 4 + C = 4, \] \[ -1 + C = 4, \] \[ C = 4 + 1 = 5. \] Thus, the \( n^\text{th} \) term is: \[ a_n = 3n^2 - 4n + 5. \] **Final Answers:** - \( x = 20 \). - The \( n^\text{th} \) term of the sequence is \( a_n = 3n^2 - 4n + 5 \).

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Bonus Knowledge

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 \( a_n = n^2 + 2n + 1 = (n + 1)^2 \).

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