QUESTION 5 A quadratic pattem has a second term equal to1, a third term equal to -6 and a fifh term equal to -14 . 5.1 Calculate the second difference of this quadratic patterm. 5.2 Hence, or otherwise, calculate the first term of the patlem.

Let the quadratic pattern be given by \[ T_n = A n^2 + B n + C. \] We are given: - \(T_2 = 4A + 2B + C = 1\), - \(T_3 = 9A + 3B + C = -6\), - \(T_5 = 25A + 5B + C = -14\). **Step 1.** Subtract the equation for \(T_2\) from the equation for \(T_3\): \[ (9A+3B+C) - (4A+2B+C) = -6 - 1. \] This simplifies to: \[ 5A + B = -7. \tag{1} \] **Step 2.** Subtract the equation for \(T_2\) from the equation for \(T_5\): \[ (25A+5B+C) - (4A+2B+C) = -14 - 1. \] This gives: \[ 21A + 3B = -15. \] Divide through by 3: \[ 7A + B = -5. \tag{2} \] **Step 3.** Subtract equation \((1)\) from equation \((2)\): \[ (7A+B) - (5A+B) = -5 - (-7). \] This yields: \[ 2A = 2 \quad \Longrightarrow \quad A = 1. \] **Step 4.** Substitute \(A=1\) into equation \((1)\): \[ 5(1) + B = -7 \quad \Longrightarrow \quad 5 + B = -7 \quad \Longrightarrow \quad B = -12. \] **Step 5.** Substitute \(A=1\) and \(B=-12\) into \(T_2\)’s equation: \[ 4(1) + 2(-12) + C = 1 \quad \Longrightarrow \quad 4 - 24 + C = 1. \] Thus: \[ -20 + C = 1 \quad \Longrightarrow \quad C = 21. \] Now the quadratic pattern is: \[ T_n = n^2 - 12n + 21. \] **5.1 Calculation of the Second Difference** For any quadratic sequence, the second difference is constant and equals \(2A\). Since \(A = 1\), the second difference is: \[ 2 \times 1 = 2. \] **5.2 Calculation of the First Term** The first term is: \[ T_1 = 1^2 - 12(1) + 21 = 1 - 12 + 21 = 10. \] Thus, the answers are: - The second difference is \(\boxed{2}\). - The first term of the pattern is \(\boxed{10}\).