(i) The sum of the first three terms of an arithmetic series is 24 and the sum of the next three terms is -12. Calculate the sum of the first twenty terms of this series.

Let's denote the first term of the arithmetic series as \( a \) and the common difference as \( d \). The first three terms of the series are: 1. \( a \) 2. \( a + d \) 3. \( a + 2d \) The sum of the first three terms can be written as: \[ S_3 = a + (a + d) + (a + 2d) = 3a + 3d = 3(a + d) \] According to the problem, this sum is 24: \[ 3(a + d) = 24 \implies a + d = 8 \tag{1} \] The next three terms of the series are: 1. \( a + 3d \) 2. \( a + 4d \) 3. \( a + 5d \) The sum of these three terms can be expressed as: \[ S_3' = (a + 3d) + (a + 4d) + (a + 5d) = 3a + 12d = 3(a + 4d) \] According to the problem, this sum is -12: \[ 3(a + 4d) = -12 \implies a + 4d = -4 \tag{2} \] Now we have two equations: 1. \( a + d = 8 \) 2. \( a + 4d = -4 \) Let's subtract equation (1) from equation (2): \[ (a + 4d) - (a + d) = -4 - 8 \] \[ 3d = -12 \implies d = -4 \] Now substituting \( d \) back into equation (1) to find \( a \): \[ a - 4 = 8 \implies a = 12 \] Thus, we have \( a = 12 \) and \( d = -4 \). Now, we can calculate the sum of the first 20 terms of the series. The formula for the sum of the first \( n \) terms \( S_n \) of an arithmetic series is: \[ S_n = \frac{n}{2} \times (2a + (n-1)d) \] Substituting \( n = 20 \), \( a = 12 \), and \( d = -4 \): \[ S_{20} = \frac{20}{2} \times (2 \cdot 12 + (20 - 1)(-4)) \] \[ = 10 \times (24 + 19 \cdot (-4)) \] \[ = 10 \times (24 - 76) \] \[ = 10 \times (-52) \] \[ = -520 \] Thus, the sum of the first twenty terms of this series is \(\boxed{-520}\).