QUESTION 3 [9] The sequence \( 3 ; 9 ; 17 ; 27 ; \ldots \) is a quadratic sequence. 3.1 Write down the next term. 3.2 Determine an expression for the \( n^{\text {th }} \) term of the sequence. (1) 3.3 What is the value of the first term of the sequence that is greater than 269 ? (4) (4)

**3.1 Next term** The given sequence is \[ 3,\ 9,\ 17,\ 27,\ \ldots \] First, compute the differences between consecutive terms: - \(9 - 3 = 6\) - \(17 - 9 = 8\) - \(27 - 17 = 10\) The differences form the sequence \[ 6,\ 8,\ 10,\ \ldots \] which increases by 2 each time. Thus, the next difference will be \[ 10 + 2 = 12. \] Therefore, the next term will be \[ 27 + 12 = 39. \] --- **3.2 Expression for the \( n^{\text{th}} \) term** Assume the \( n^{\text{th}} \) term of a quadratic sequence is given by \[ a_n = an^2 + bn + c. \] Using the first three terms, we set up the equations: For \( n = 1 \): \[ a(1)^2 + b(1) + c = a + b + c = 3. \] For \( n = 2 \): \[ a(2)^2 + b(2) + c = 4a + 2b + c = 9. \] For \( n = 3 \): \[ a(3)^2 + b(3) + c = 9a + 3b + c = 17. \] **Step 1. Find \( a \) and \( b \)** Subtract the \( n = 1 \) equation from the \( n = 2 \) equation: \[ (4a + 2b + c) - (a + b + c) = 9 - 3, \] which simplifies to: \[ 3a + b = 6. \quad \text{(Equation 1)} \] Subtract the \( n = 2 \) equation from the \( n = 3 \) equation: \[ (9a + 3b + c) - (4a + 2b + c) = 17 - 9, \] which simplifies to: \[ 5a + b = 8. \quad \text{(Equation 2)} \] Subtract Equation 1 from Equation 2: \[ (5a + b) - (3a + b) = 8 - 6, \] which gives: \[ 2a = 2 \quad \Rightarrow \quad a = 1. \] Substitute \( a = 1 \) into Equation 1: \[ 3(1) + b = 6 \quad \Rightarrow \quad 3 + b = 6 \quad \Rightarrow \quad b = 3. \] **Step 2. Find \( c \)** Substitute \( a = 1 \) and \( b = 3 \) into the equation for \( n = 1 \): \[ 1 + 3 + c = 3 \quad \Rightarrow \quad 4 + c = 3 \quad \Rightarrow \quad c = -1. \] Thus, the \( n^{\text{th}} \) term is given by: \[ \boxed{a_n = n^2 + 3n - 1}. \] --- **3.3 First term greater than 269** We need to find the smallest \( n \) such that \[ a_n > 269, \] where \[ a_n = n^2 + 3n - 1. \] Set up the equation: \[ n^2 + 3n - 1 = 269. \] Simplify: \[ n^2 + 3n - 270 = 0. \] Solve the quadratic equation: \[ n = \frac{-3 \pm \sqrt{3^2 - 4(1)(-270)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 1080}}{2} = \frac{-3 \pm \sqrt{1089}}{2}. \] Since \(\sqrt{1089} = 33\), we have: \[ n = \frac{-3 \pm 33}{2}. \] The positive solution is: \[ n = \frac{-3 + 33}{2} = \frac{30}{2} = 15. \] This means \[ a_{15} = 15^2 + 3(15) - 1 = 225 + 45 - 1 = 269. \] Since we need the first term **greater than** 269, we take \( n = 16 \): \[ a_{16} = 16^2 + 3(16) - 1 = 256 + 48 - 1 = 303. \] Thus, the first term greater than 269 is: \[ \boxed{303}. \]