2.1 Given a quadratic number pattern: \( -120 ;-99 ;-80 ;-63 ; \ldots \) 2.1.1 Write down the next TWO terms of the pattern. 2.1.2 Determine the \( \mathrm{n}^{\text {th }} \) term of the number pattern in the form \( \mathrm{T}_{\mathrm{n}}=\mathrm{an}^{2}+\mathrm{bn}+\mathrm{c} \). 2.1.3 What value must be added to \( \mathrm{T}_{\mathrm{n}} \) for the sequence to have only one value of n for which \( \mathrm{T}_{\mathrm{n}}=0 \) ?

To solve the problem step by step, let's start with the given quadratic number pattern: \( -120, -99, -80, -63, \ldots \). ### 2.1.1 Write down the next TWO terms of the pattern. First, we need to find the differences between the consecutive terms: - The difference between the first and second terms: \[ -99 - (-120) = 21 \] - The difference between the second and third terms: \[ -80 - (-99) = 19 \] - The difference between the third and fourth terms: \[ -63 - (-80) = 17 \] Now, we can see that the differences are decreasing by 2 each time: - First difference: \( 21 \) - Second difference: \( 19 \) - Third difference: \( 17 \) Following this pattern, the next difference should be \( 15 \) and then \( 13 \). Now, we can find the next two terms: - The fifth term: \[ -63 + 15 = -48 \] - The sixth term: \[ -48 + 13 = -35 \] Thus, the next two terms are \( -48 \) and \( -35 \). ### 2.1.2 Determine the \( \mathrm{n}^{\text {th }} \) term of the number pattern in the form \( \mathrm{T}_{\mathrm{n}}=\mathrm{an}^{2}+\mathrm{bn}+\mathrm{c} \). To find the \( n^{th} \) term, we can use the known terms to set up a system of equations. We have: - \( T_1 = -120 \) - \( T_2 = -99 \) - \( T_3 = -80 \) - \( T_4 = -63 \) This gives us the following equations: 1. \( a(1^2) + b(1) + c = -120 \) → \( a + b + c = -120 \) (1) 2. \( a(2^2) + b(2) + c = -99 \) → \( 4a + 2b + c = -99 \) (2) 3. \( a(3^2) + b(3) + c = -80 \) → \( 9a + 3b + c = -80 \) (3) Now, we can solve this system of equations. Subtract equation (1) from equation (2): \[ (4a + 2b + c) - (a + b + c) = -99 + 120 \] \[ 3a + b = 21 \quad (4) \] Subtract equation (2) from equation (3): \[ (9a + 3b + c) - (4a + 2b + c) = -80 + 99 \] \[ 5a + b = 19 \quad (5) \] Now, we can subtract equation (4) from equation (5): \[ (5a + b) - (3a + b) = 19 - 21 \] \[ 2a = -2 \implies a = -1 \] Now substitute \( a = -1 \) back into equation (4): \[ 3(-1) + b = 21 \implies -3 + b = 21 \implies b = 24 \] Now substitute \( a = -1 \) and \( b = 24 \) back into equation (1): \[ -1 + 24 + c = -120 \implies 23 + c = -120 \implies c = -143 \] Thus, the \( n^{th} \) term is: \[ T_n = -n^2 + 24n - 143 \] ### 2.1.3 What value must be added to \( T_n \) for the sequence to have only one value of \( n \) for which \( T_n = 0 \)? For the quadratic equation \( T_n = -n^2 + 24n - 143 + k = 0 \) to have only one solution, the discriminant must be zero. The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ D = b^2 - 4ac \] In our case: - \( a = -1 \) - \( b = 24 \) - \( c = -143 + k \) Setting the discriminant to zero: \[ D = 24^2 - 4(-1)(-143 + k) = 0 \] \[ 576 - 4(143 - k) = 0 \] \[ 576 - 572 + 4k = 0 \] \[ 4k = -4 \implies k = -1 \] Thus, the value that must be added to \( T_n \) for the sequence to have only one value of \( n \) for which \( T_n = 0 \) is \( -1 \). ### Summary of Answers: - **2.1.1**: The next two terms are \( -48 \) and \( -35 \). - **2.1.2**: The \( n^{th} \) term is \( T_n = -n^2 + 24n - 143 \). - **2.1.3**: The value to be added is \( -1 \).