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.

We start with the quadratic sequence given by \[ T_{n}=a n^{2}+b n+6. \] The first term is \[ T_{1}=a+b+6, \] the second term is \[ T_{2}=4a+2b+6, \] and the third term is \[ T_{3}=9a+3b+6. \] **Step 1. Use the first condition.** We are told that the second term is 15 more than the first term: \[ T_{2}=T_{1}+15. \] Substituting the expressions for \(T_1\) and \(T_2\): \[ 4a+2b+6 = (a+b+6)+15. \] Simplify the right-hand side: \[ 4a+2b+6 = a+b+21. \] Subtract \(a+b+6\) from both sides: \[ (4a - a) + (2b - b) = 21 - 6, \] \[ 3a+b=15. \tag{1} \] **Step 2. Use the second condition.** The problem states that the third term is comprised of the second term and 10 more than the first term. This gives: \[ T_{3}=T_{2}+(T_{1}+10). \] Substitute the expressions for \(T_1\), \(T_2\), and \(T_3\): \[ 9a+3b+6 = (4a+2b+6) + (a+b+6+10). \] Simplify the right-hand side: \[ 9a+3b+6 = (4a+2b+6) + (a+b+16) = (4a+a)+(2b+b)+(6+16), \] \[ 9a+3b+6 = 5a+3b+22. \] Eliminate \(3b\) from both sides: \[ 9a+6=5a+22. \] Solve for \(a\) by subtracting \(5a\) and then \(6\) from both sides: \[ 4a =16 \quad \Longrightarrow \quad a=4. \] **Step 3. Find \(b\).** Substitute \(a=4\) into equation \((1)\): \[ 3(4)+b=15, \] \[ 12+b=15 \quad \Longrightarrow \quad b=3. \] **Step 4. Determine \(T_{1}\) and \(T_{3}\) and compute their sum.** Calculate the first term \(T_{1}\): \[ T_{1}=a+b+6=4+3+6=13. \] Calculate the third term \(T_{3}\): \[ T_{3}=9a+3b+6=9(4)+3(3)+6=36+9+6=51. \] Thus, the sum of \(T_{3}\) and \(T_{1}\) is: \[ T_{3}+T_{1}=51+13=64. \] \[ \boxed{64} \]