9.1 The sequence \( 4 ; 9 ; x ; 37 ; \ldots j \) a quadratic sequence. 9.1 Calculate \( x \). 9.2 Hence, or, otherwise, determine the nth term of the sequence 9.2 Consider the following quadratic sequence: \( 6 ; x ; 26 ; 45 ; y ; \)... Calculate the values of \( x \) and \( y \). 9.3 Given the following series: \( -11-4+3+\ldots+200 \) 9.2.1 Calculate the sum of the series. 9.2 .2 Write the series in sigma notation.

Let's solve the problems step by step. ### 9.1 Calculate \( x \) in the sequence \( 4, 9, x, 37, \ldots, j \) A quadratic sequence can be defined by a second difference that is constant. Let's denote the terms of the sequence as follows: - \( a_1 = 4 \) - \( a_2 = 9 \) - \( a_3 = x \) - \( a_4 = 37 \) First, we calculate the first differences: - \( a_2 - a_1 = 9 - 4 = 5 \) - \( a_3 - a_2 = x - 9 \) - \( a_4 - a_3 = 37 - x \) Now, we calculate the second differences: - First difference between the first two first differences: \[ (x - 9) - 5 = x - 14 \] - Second difference between the last two first differences: \[ (37 - x) - (x - 9) = 37 - x - x + 9 = 46 - 2x \] Since the second differences must be equal, we set them equal to each other: \[ x - 14 = 46 - 2x \] Now, let's solve for \( x \): \[ x + 2x = 46 + 14 \] \[ 3x = 60 \] \[ x = 20 \] ### 9.2 Hence, or otherwise, determine the nth term of the sequence Now that we have \( x = 20 \), the sequence is \( 4, 9, 20, 37, \ldots \). To find the nth term, we can use the general form of a quadratic sequence: \[ a_n = An^2 + Bn + C \] We can set up a system of equations using the known terms: 1. For \( n = 1 \): \( A(1^2) + B(1) + C = 4 \) → \( A + B + C = 4 \) 2. For \( n = 2 \): \( A(2^2) + B(2) + C = 9 \) → \( 4A + 2B + C = 9 \) 3. For \( n = 3 \): \( A(3^2) + B(3) + C = 20 \) → \( 9A + 3B + C = 20 \) Now we can solve this system of equations. Let's denote the equations as follows: 1. \( A + B + C = 4 \) (Equation 1) 2. \( 4A + 2B + C = 9 \) (Equation 2) 3. \( 9A + 3B + C = 20 \) (Equation 3) We can eliminate \( C \) by subtracting Equation 1 from Equations 2 and 3. Subtracting Equation 1 from Equation 2: \[ (4A + 2B + C) - (A + B + C) = 9 - 4 \] \[ 3A + B = 5 \quad (Equation 4) \] Subtracting Equation 1 from Equation 3: \[ (9A + 3B + C) - (A + B + C) = 20 - 4 \] \[ 8A + 2B = 16 \quad (Equation 5) \] Now we can simplify Equation 5: \[ 4A + B = 8 \quad (Equation 6) \] Now we have a new system of equations: 1. \( 3A + B = 5 \) (Equation 4) 2. \( 4A + B = 8 \) (Equation 6) Subtract Equation 4 from Equation 6: \[ (4A + B) - (3A + B) = 8 - 5 \] \[ A = 3 \] Substituting \( A = 3 \) back into Equation 4: \[ 3(3) + B = 5 \] \[ 9 + B = 5 \] \[ B = -4 \] Now substituting \( A \) and \( B \) back into Equation 1 to find \( C \): \[ 3 - 4 + C = 4 \] \[ C = 5 \] Thus, the nth term of the sequence is: \[ a_n = 3n^2 - 4n + 5 \] ### 9.2 Calculate the values of \( x \) and \( y \) in the sequence \( 6, x, 26, 45, y \) Using the same method as before, we denote the terms: - \( b_1 = 6 \) - \( b_2 = x \) - \( b_3 = 26 \) - \( b_4 = 45 \) - \( b_5 = y \) Calculating the first differences: - \( b_2 - b_1 = x - 6 \) - \( b_3 - b_2 = 26 - x \) - \( b_4 - b_3 = 45 - 26 = 19 \) Now, we calculate the second differences: - First difference between the first two first differences: \[ (26 - x) - (x - 6) = 26 - x - x + 6 = 32 - 2x \] - Second difference between the last two first differences: \[ 19 - (26 - x) = 19 - 26 + x = x - 7 \] Setting the second differences equal: \[ 32 - 2x = x - 7 \] Now, let's solve for \( x \): \[ 32 + 7 = 3x \] \[ 39 = 3x \] \[ x = 13 \] Now we have \( x = 13 \). The sequence is \( 6, 13, 26, 45, y \). Next, we can find \( y \) by calculating the first differences again: - \( b_1 = 6 \) - \( b_2 = 13 \) - \( b_3 = 26 \) - \( b_4 = 45 \) - \( b_5 = y \) Calculating the first differences: - \( b_2 - b_1 = 13 - 6 = 7 \) - \( b_3 - b_2 = 26 - 13 = 13 \) - \( b_4 - b_3 = 45 - 26 = 19 \) - \( b_5 - b_4 = y - 45 \