(a) The first five terms of a sequence are \( 8,3 / 5,2,-1 \) and -4 . (i) Find an expression, in terms of \( n \), for the \( n \)th term of the sequence. Answer ..............

To find an expression for the \( n \)th term of the sequence given the first five terms \( 8, \frac{3}{5}, 2, -1, -4 \), we first observe the pattern in the sequence. Let's denote the terms as follows: - \( a_1 = 8 \) - \( a_2 = \frac{3}{5} \) - \( a_3 = 2 \) - \( a_4 = -1 \) - \( a_5 = -4 \) Next, we can calculate the differences between consecutive terms to see if there is a pattern: 1. \( a_2 - a_1 = \frac{3}{5} - 8 = \frac{3 - 40}{5} = \frac{-37}{5} \) 2. \( a_3 - a_2 = 2 - \frac{3}{5} = \frac{10 - 3}{5} = \frac{7}{5} \) 3. \( a_4 - a_3 = -1 - 2 = -3 \) 4. \( a_5 - a_4 = -4 - (-1) = -3 \) Now, let's look at the second differences: 1. \( \frac{7}{5} - \frac{-37}{5} = \frac{7 + 37}{5} = \frac{44}{5} \) 2. \( -3 - \frac{7}{5} = -3 - \frac{7}{5} = \frac{-15 - 7}{5} = \frac{-22}{5} \) 3. \( -3 - (-3) = 0 \) The first differences do not seem to form a clear pattern, but the second differences suggest that the sequence might be quadratic. To find a quadratic expression, we can assume the \( n \)th term can be expressed as: \[ a_n = An^2 + Bn + C \] We can set up a system of equations using the first three terms: 1. For \( n = 1 \): \[ A(1^2) + B(1) + C = 8 \quad \Rightarrow \quad A + B + C = 8 \quad \text{(1)} \] 2. For \( n = 2 \): \[ A(2^2) + B(2) + C = \frac{3}{5} \quad \Rightarrow \quad 4A + 2B + C = \frac{3}{5} \quad \text{(2)} \] 3. For \( n = 3 \): \[ A(3^2) + B(3) + C = 2 \quad \Rightarrow \quad 9A + 3B + C = 2 \quad \text{(3)} \] Now we have the following system of equations: 1. \( A + B + C = 8 \) 2. \( 4A + 2B + C = \frac{3}{5} \) 3. \( 9A + 3B + C = 2 \) Next, we can eliminate \( C \) by subtracting equations: From (2) - (1): \[ (4A + 2B + C) - (A + B + C) = \frac{3}{5} - 8 \] \[ 3A + B = \frac{3}{5} - \frac{40}{5} = \frac{-37}{5} \quad \text{(4)} \] From (3) - (2): \[ (9A + 3B + C) - (4A + 2B + C) = 2 - \frac{3}{5} \] \[ 5A + B = 2 - \frac{3}{5} = \frac{10}{5} - \frac{3}{5} = \frac{7}{5} \quad \text{(5)} \] Now we have a new system of equations: 1. \( 3A + B = \frac{-37}{5} \) (4) 2. \( 5A + B = \frac{7}{5} \) (5) Subtract (4) from (5): \[ (5A + B) - (3A + B) = \frac{7}{5} - \frac{-37}{5} \] \[ 2A = \frac{7 + 37}{5} = \frac{44}{5} \] \[ A = \