1.4 A quadratic pattern has a second term equal to 1 and a third term equal to -6 and a fifth term equal to -14 . 1.4 .1 calculate the second differences of this quadratic pattern 1.4 .2 hence, or otherwise, calculate the first term of the pattern 1.5 \( \begin{array}{l}\text { The sequence } 4 ; 9 \text {; The pattern } 4 ; 9 ; x, 37 ; \ldots \text { is a quadratic pattern } \\ 1.5 .1 \text { Calculate the value of } x \text {. } \\ 1.5 .2 \text { (3) } \\ \text { hence or otherwise determine the general term of this pattern. } \\ 1.6\end{array} \begin{array}{l}\text { Consider the sequence: } 3 ; p ; 10 ; q ; 21 \text {. The sequence has a constant second } \\ \text { difference of 1. Determine the values of } p \text { and } q \text {. }\end{array} \)

Let's break down the problem step by step. ### 1.4 Quadratic Pattern We know that a quadratic pattern can be expressed in the form: \[ a_n = An^2 + Bn + C \] where \(A\), \(B\), and \(C\) are constants, and \(n\) is the term number. #### 1.4.1 Calculate the second differences Given: - Second term (\(a_2\)) = 1 - Third term (\(a_3\)) = -6 - Fifth term (\(a_5\)) = -14 We can express these terms using the quadratic formula: - \(a_2 = 4A + 2B + C = 1\) - \(a_3 = 9A + 3B + C = -6\) - \(a_5 = 25A + 5B + C = -14\) We can set up the following system of equations: 1. \(4A + 2B + C = 1\) (Equation 1) 2. \(9A + 3B + C = -6\) (Equation 2) 3. \(25A + 5B + C = -14\) (Equation 3) Next, we will solve this system of equations to find \(A\), \(B\), and \(C\). #### 1.4.2 Calculate the first term of the pattern The first term \(a_1\) can be calculated using the values of \(A\), \(B\), and \(C\). Let's solve the system of equations. ### Solving the System of Equations We will subtract Equation 1 from Equation 2 and Equation 2 from Equation 3 to eliminate \(C\). 1. Subtract Equation 1 from Equation 2: \[ (9A + 3B + C) - (4A + 2B + C) = -6 - 1 \] This simplifies to: \[ 5A + B = -7 \quad \text{(Equation 4)} \] 2. Subtract Equation 2 from Equation 3: \[ (25A + 5B + C) - (9A + 3B + C) = -14 + 6 \] This simplifies to: \[ 16A + 2B = -8 \quad \text{(Equation 5)} \] Now we can solve Equations 4 and 5 together. ### Solving Equations 4 and 5 From Equation 4: \[ B = -7 - 5A \] Substituting \(B\) into Equation 5: \[ 16A + 2(-7 - 5A) = -8 \] This simplifies to: \[ 16A - 14 - 10A = -8 \] \[ 6A - 14 = -8 \] \[ 6A = 6 \] \[ A = 1 \] Now substituting \(A\) back into Equation 4 to find \(B\): \[ 5(1) + B = -7 \] \[ B = -12 \] Now substituting \(A\) and \(B\) back into Equation 1 to find \(C\): \[ 4(1) + 2(-12) + C = 1 \] \[ 4 - 24 + C = 1 \] \[ C = 21 \] Now we have: - \(A = 1\) - \(B = -12\) - \(C = 21\) The first term \(a_1\) can be calculated as: \[ a_1 = A(1^2) + B(1) + C = 1 - 12 + 21 = 10 \] ### 1.5 Quadratic Pattern Given the sequence \(4, 9, x, 37, \ldots\), we know that it is a quadratic pattern. #### 1.5.1 Calculate the value of \(x\) We can express the terms as: - \(a_1 = 4\) - \(a_2 = 9\) - \(a_3 = x\) - \(a_4 = 37\) We can find the second differences to determine \(x\). The first differences are: - \(d_1 = 9 - 4 = 5\) - \(d_2 = x - 9\) - \(d_3 = 37 - x\) The second differences are: - \(d_2 - d_1 = (x - 9) - 5 = x - 14\) - \(d_3 - d_2 = (37 - x) - (x - 9) = 46 - 2x\) Since the second differences are constant, we can set them equal: \[ x - 14 = 46 - 2x \] Solving for \(x\): \[ 3x = 60 \] \[ x = 20 \] #### 1.5.2 Determine the general term of this pattern Now we have: - \(a_1 = 4\) - \(a_2 = 9\) - \(a_3 = 20\) - \(a_4 = 37\) We can find the general term \(a_n = An^2 + Bn + C\) using the known terms. Using the values of \(A\), \(B\), and \(C\) from the previous calculations, we can find the general term. ### 1.6 Sequence with Constant Second Difference Given the sequence \(3, p, 10, q, 21\) with a constant second difference of 1. The first differences are: - \(d_1 = p - 3\) - \(d_2 = 10 - p\) - \(d_3 = q - 10\) - \(d_4 = 21 - q\) The second differences are: - \(d_2 - d_1 = (10 - p) - (p - 3) = 13 - 2p\) - \(d_3 - d_2 = (q - 10) - (10 - p) = q - 20 + p\) - \(d_4 - d_3 = (21 - q) - (q - 10) = 31 - 2q\) Since the second differences are constant and equal to 1, we can set up the equations: 1. \(13 - 2p = 1\) 2. \(q - 20 + p = 1\) 3. \(31 - 2q = 1\) Let's solve these equations to find \(p\) and \(q\). ### Solving for \(p\) and \(q\) 1. From \(13