6, 16, 30,48,70, The \( n \)th term of this sequence is given by \( T_{n}=a n+b n^{2} \) where \( a \) and \( b \) are numbers. a By looking at the first term, show that \( a+b=6 \). b Use the second term to find another equation involving \( a \) and \( b \). c Solve the equation from parts a and \( \mathbf{b} \) simultaneously to find the \( n \)th term of the sequence.

To solve the problem, we will analyze the sequence and derive the equations step by step. ### Part a: Show that \( a + b = 6 \) The \( n \)th term of the sequence is given by: \[ T_n = a n + b n^2 \] For the first term (\( n = 1 \)), we have: \[ T_1 = a(1) + b(1^2) = a + b \] From the sequence, we know that \( T_1 = 6 \). Therefore, we can write: \[ a + b = 6 \quad \text{(Equation 1)} \] ### Part b: Use the second term to find another equation involving \( a \) and \( b \) For the second term (\( n = 2 \)), we have: \[ T_2 = a(2) + b(2^2) = 2a + 4b \] From the sequence, we know that \( T_2 = 16 \). Therefore, we can write: \[ 2a + 4b = 16 \quad \text{(Equation 2)} \] ### Part c: Solve the equations from parts a and b simultaneously We have the following system of equations: 1. \( a + b = 6 \) (Equation 1) 2. \( 2a + 4b = 16 \) (Equation 2) We can simplify Equation 2 by dividing everything by 2: \[ a + 2b = 8 \quad \text{(Equation 3)} \] Now we can solve Equations 1 and 3 simultaneously. We have: 1. \( a + b = 6 \) 2. \( a + 2b = 8 \) Subtract Equation 1 from Equation 3: \[ (a + 2b) - (a + b) = 8 - 6 \] This simplifies to: \[ b = 2 \] Now, substitute \( b = 2 \) back into Equation 1: \[ a + 2 = 6 \] This gives us: \[ a = 4 \] ### Conclusion We have found \( a = 4 \) and \( b = 2 \). Therefore, the \( n \)th term of the sequence is: \[ T_n = 4n + 2n^2 \] We can also write it in standard form: \[ T_n = 2n^2 + 4n \] Thus, the \( n \)th term of the sequence is: \[ T_n = 2n^2 + 4n \]