(o) A certain sequence has a constant second difference of 8. The sixth term of the sequence is 172 and the tenth term is 452. Determine the general term.

To find the general term of the sequence with a constant second difference of 8, we begin with the understanding that this implies the sequence is quadratic in nature, i.e., of the form \( a_n = An^2 + Bn + C \). Given: - \( a_6 = 172 \) - \( a_{10} = 452 \) We can derive two equations based on these given terms. 1. For \( n = 6 \): \[ 36A + 6B + C = 172 \quad \text{(Equation 1)} \] 2. For \( n = 10 \): \[ 100A + 10B + C = 452 \quad \text{(Equation 2)} \] Next, we need another equation. Since the second difference is constant, we can express this condition in terms of the coefficients: The second difference is \( 2A \). Given that the second difference is 8, we have: \[ 2A = 8 \implies A = 4 \] With \( A = 4 \), we can substitute \( A \) back into our equations. Substituting \( A = 4 \) into Equation 1 and Equation 2, we get: **From Equation 1:** \[ 36(4) + 6B + C = 172 \implies 144 + 6B + C = 172 \implies 6B + C = 28 \quad \text{(Equation 3)} \] **From Equation 2:** \[ 100(4) + 10B + C = 452 \implies 400 + 10B + C = 452 \implies 10B + C = 52 \quad \text{(Equation 4)} \] Now we have two equations (Equation 3 and Equation 4): 1. \( 6B + C = 28 \) 2. \( 10B + C = 52 \) To eliminate \( C \), we can subtract Equation 3 from Equation 4: \[ (10B + C) - (6B + C) = 52 - 28 \implies 4B = 24 \implies B = 6 \] Now substituting \( B = 6 \) back into Equation 3: \[ 6(6) + C = 28 \implies 36 + C = 28 \implies C = 28 - 36 \implies C = -8 \] Now we have all coefficients: - \( A = 4 \) - \( B = 6 \) - \( C = -8 \) The general term of the sequence is: \[ a_n = 4n^2 + 6n - 8 \] Thus, the general term of the sequence is: \[ \boxed{4n^2 + 6n - 8} \]