A quadratic number pattern has a general term \( T_{n}=a n^{2}+b n+c . T_{2}-T_{1}=4 \) and \( T_{3}-T_{2}=-2 \). The third term of the sequence is 7 . Determine the values of \( a, b \) and

To solve for \( a \), \( b \), and \( c \), we start with the general term of the quadratic sequence given by \( T_n = a n^2 + b n + c \). From the problem, we know: 1. \( T_2 - T_1 = 4 \) 2. \( T_3 - T_2 = -2 \) 3. \( T_3 = 7 \) First, let's express \( T_1 \), \( T_2 \), and \( T_3 \): - \( T_1 = a(1^2) + b(1) + c = a + b + c \) - \( T_2 = a(2^2) + b(2) + c = 4a + 2b + c \) - \( T_3 = a(3^2) + b(3) + c = 9a + 3b + c \) Using the first equation \( T_2 - T_1 = 4 \): \[ (4a + 2b + c) - (a + b + c) = 4 \\ 4a + 2b + c - a - b - c = 4 \\ 3a + b = 4 \quad \text{(1)} \] Now, using the second equation \( T_3 - T_2 = -2 \): \[ (9a + 3b + c) - (4a + 2b + c) = -2 \\ 9a + 3b + c - 4a - 2b - c = -2 \\ 5a + b = -2 \quad \text{(2)} \] Next, we can solve this system of equations: From (1), we have: \[ b = 4 - 3a \quad \text{(3)} \] Substituting (3) into (2): \[ 5a + (4 - 3a) = -2 \\ 5a + 4 - 3a = -2 \\ 2a + 4 = -2 \\ 2a = -6 \\ a = -3 \] Now substituting back \( a = -3 \) into (3) to find \( b \): \[ b = 4 - 3(-3) \\ b = 4 + 9 = 13 \] Next, we use the value of \( a \) and \( b \) to find \( c \) using the third term \( T_3 = 7 \): \[ T_3 = 9a + 3b + c = 7 \\ 9(-3) + 3(13) + c = 7 \\ -27 + 39 + c = 7 \\ 12 + c = 7 \\ c = 7 - 12 = -5 \] Finally, we have: \[ a = -3, \quad b = 13, \quad c = -5 \] Thus, the values of \( a \), \( b \), and \( c \) are \( a = -3 \), \( b = 13 \), and \( c = -5 \).