5.1 The \( p^{\text {th }} \) term of the first difference of a quadratic sequence is given by \( T_{p}=3 p-2 \) \( 51.1 \quad \begin{array}{l}\text { Determine between which } \\ \text { two consecutive terms of the } \\ \text { quadratic sequence the first } \\ \text { difference is equal to } 1450\end{array} \) \( \begin{array}{l}\text { (4) } \mathbf{L . 3}\end{array} \) \( \begin{array}{l}\text { If } T_{40}=2290 \text { and } T_{n}=a n^{2}+b n+c \\ n^{\text {th }} \text { term of the quadratic sequence } \\ \text { Calculate the value of } \mathrm{c}\end{array} \)

To find out between which two consecutive terms of the quadratic sequence the first difference equals 1450, we can set \( T_{p} = 1450 \). From the first difference formula: \[ 3p - 2 = 1450 \] Solving for \( p \): \[ 3p = 1452 \\ p = 484 \] This means that the first difference is equal to 1450 between the \( 484^{\text{th}} \) term and the \( 485^{\text{th}} \) term of the quadratic sequence. To find the value of \( c \) in the quadratic sequence given that \( T_{40} = 2290 \) and is represented by \( T_{n} = an^{2} + bn + c \), we also need to express \( T_{39} \) and \( T_{38} \) using the first difference: The first difference for \( p = 39 \) is: \[ T_{39} = 3(39) - 2 = 115 \] Then we calculate \( T_{39} \): \[ T_{39} = T_{40} - T_{39} + 1450 \] As \( T_{40} = 2290 \), \[ T_{39} = 2290 - 115 = 2175 \] Now we can express: \[ T_{40} = 40a + 40b + c = 2290 \\ T_{39} = 39a + 39b + c = 2175 \] Subtracting these equations will help eliminate \( c \): \[ (40a + 40b + c) - (39a + 39b + c) = 2290 - 2175 \\ a + b = 115 \] From here, you could further explore and substitute back to find \( c \). Remember that the difficulty lies in solving these equations, but keep focused on isolating \( c \) by determining the values of \( a \) and \( b \). Happy solving!