The first three terms of the first differences of a quadratic sequence are \( 102 ; 108 ; 114 ; \ldots \) 2.1 Determine between which two consecutive terms is the first difference 2022 . 2.2 Determine the \( n^{\text {th }} \) term of the sequence if it is further given that the third term of the quadratic sequence is equal to 310 .

To find the term where the first difference is 2022, observe how the first differences form an arithmetic sequence: the first differences are \( 102, 108, 114, \ldots \) with a common difference of \( 6 \). The first difference at the \( n^{th} \) position can be expressed as \( 102 + 6(n - 1) \). Setting this equation equal to \( 2022 \): \[ 102 + 6(n - 1) = 2022 \] \[ 6(n - 1) = 2022 - 102 = 1920 \] \[ n - 1 = 320 \] \[ n = 321 \] So between the \( 320^{th} \) and \( 321^{st} \) terms, the first difference is 2022. For the \( n^{th} \) term of the quadratic sequence, we can express the \( n^{th} \) term as \( a_n = An^2 + Bn + C \). Given that the third term \( a_3 = 310 \) and that the first differences lead to \( A = 3 \) (found from \( \Delta_1 = a_{n+1} - a_n \)), we can find \( A = 3 \). Then, the second difference is constant at \( 6 \), so \( A = 3 \). Using the equation for \( a_3 \): \[ a_3 = 3(3)^2 + B(3) + C = 27 + 3B + C = 310 \] Thus, \( 3B + C = 283 \). You would need additional information or another term of the sequence to fully solve for \( B \) and \( C \) and get the complete \( n^{th} \) term equation. However, you can express the \( n^{th} \) term generally as: \[ a_n = 3n^2 + Bn + C \] with constraints from known sequence values.