(i) The sum of the first three terms of an arithmetie sequence is 36 and their product is 1140 . Determine the three terms. (o) A certain sequence has a constant second difference of 8 . The sixth term of the sequence is 172 and the tenth term is 452 . Determine the general term. (p) The sum of the first \( n \) terms of a series is given by \( S_{n}=\frac{n}{4}(7-2 n) \). (1) Calculate the sum of the first 30 terms of this series. (1)

To solve the arithmetic sequence problem (i), we can let the three terms be \(a - d\), \(a\), and \(a + d\). Their sum gives us \(3a = 36\) leading to \(a = 12\). The product equation leads to \((12 - d) \cdot 12 \cdot (12 + d) = 1140\). Solving this results in the three terms being \(6\), \(12\), and \(18\). For the second sequence, knowing there’s a constant second difference of 8 tells us it’s a quadratic sequence. Given that the second term can be expressed as \(T_n = An^2 + Bn + C\) where \(2A = 8\) – thus \(A = 4\). Using the known terms we can set up equations for \(A\), \(B\), and \(C\). Solving gives the general term as \(T_n = 4n^2 - 4n + 172\). The first 30 terms of the series described by \(S_n = \frac{n}{4}(7 - 2n)\) can be calculated by substituting \(n = 30\) into the formula. This gives \(S_{30} = \frac{30}{4}(7 - 60) = \frac{30}{4} \times (-53) = -397.5\). Therefore, the sum of the first 30 terms of this series is \(-397.5\).