Given a quadratic sequence \( T_{n}=a n^{2}+b n+6 \). The first term is unknown. The \( 2^{\text {nd }} \) term is 15 more than the first and the \( 3^{\text {rd }} \) term is comprised of the \( 2^{\text {nd }} \) term and 10 more than the \( 1^{\text {st }} \) term. Determine the sum of the \( 3^{\text {rd }} \) and \( 1^{\text {st }} \) term.

Let's denote the first term of the sequence as \( T_1 \). Given the sequence formula \( T_n = an^2 + bn + 6 \): 1. The first term \( T_1 \) is: \[ T_1 = a(1)^2 + b(1) + 6 = a + b + 6 \] 2. The second term \( T_2 \) is: \[ T_2 = a(2)^2 + b(2) + 6 = 4a + 2b + 6 \] According to the problem, \( T_2 \) is 15 more than the first term: \[ T_2 = T_1 + 15 \] This gives us the equation: \[ 4a + 2b + 6 = a + b + 6 + 15 \] Simplifying this equation: \[ 4a + 2b + 6 = a + b + 21 \implies 3a + b - 15 = 0 \implies b = 15 - 3a \quad (1) \] 3. The third term \( T_3 \) is: \[ T_3 = a(3)^2 + b(3) + 6 = 9a + 3b + 6 \] According to the problem, \( T_3 \) is \( T_2 + 10 \) more than the first term: \[ T_3 = T_2 + T_1 + 10 \] Substituting \( T_2 \) and \( T_1 \): \[ 9a + 3b + 6 = (4a + 2b + 6) + (a + b + 6) + 10 \] This simplifies to: \[ 9a + 3b + 6 = 5a + 3b + 22 \implies 4a - 16 = 0 \implies a = 4 \quad (2) \] 4. Substitute \( a = 4 \) back into equation (1): \[ b = 15 - 3(4) = 15 - 12 = 3 \] 5. Now, we can find \( T_1, T_2, \) and \( T_3 \): - \( T_1 = 4 + 3 + 6 = 13 \) - \( T_2 = 4(2^2) + 3(2) + 6 = 16 + 6 + 6 = 28 \) - \( T_3 = 4(3^2) + 3(3) + 6 = 36 + 9 + 6 = 51 \) 6. Finally, we need to determine the sum of the \( 3^{rd} \) and \( 1^{st} \) terms: \[ T_3 + T_1 = 51 + 13 = 64 \] Thus, the sum of the \( 3^{rd} \) and \( 1^{st} \) term is \( \boxed{64} \).