1. **Problem a(i):** Produce a sequence for the first four palings where the first paling is 1760 mm and each successive paling is 14 mm higher. The first term $a_1 = 1760$ mm. The common difference $d = 14$ mm. The sequence is arithmetic: $a_n = a_1 + (n-1)d$. Calculate the first four terms: $$a_1 = 1760$$ $$a_2 = 1760 + 14 = 1774$$ $$a_3 = 1760 + 2 \times 14 = 1788$$ $$a_4 = 1760 + 3 \times 14 = 1802$$ So, the first four palings are 1760 mm, 1774 mm, 1788 mm, and 1802 mm. 2. **Problem a(ii):** Write an expression for the height of the $n^{th}$ paling. Since this is an arithmetic sequence: $$a_n = 1760 + (n-1) \times 14$$ This formula gives the height of the $n^{th}$ paling. 3. **Problem a(iii):** Calculate the height of the last paling (the 17th paling). Using the formula: $$a_{17} = 1760 + (17-1) \times 14 = 1760 + 16 \times 14 = 1760 + 224 = 1984$$ So, the height of the 17th paling is 1984 mm. 4. **Problem b:** The first three terms of a geometric sequence are positive numbers with sum 42 and common ratio $\frac{1}{2}$. Find the first three terms. Let the first term be $a$. The terms are: $$a, a \times \frac{1}{2}, a \times \left(\frac{1}{2}\right)^2 = a, \frac{a}{2}, \frac{a}{4}$$ Sum is 42: $$a + \frac{a}{2} + \frac{a}{4} = 42$$ Combine terms: $$\frac{4a}{4} + \frac{2a}{4} + \frac{a}{4} = 42$$ $$\frac{7a}{4} = 42$$ Solve for $a$: $$a = \frac{42 \times 4}{7} = 24$$ So the terms are: $$24, 12, 6$$ 5. **Problem c:** A geometric series has first term 100 and sum to infinity 200. Find the second term. Sum to infinity formula for geometric series with $|r|<1$: $$S_\infty = \frac{a}{1-r}$$ Given: $$100 / (1-r) = 200$$ Solve for $r$: $$1-r = \frac{100}{200} = \frac{1}{2}$$ $$r = 1 - \frac{1}{2} = \frac{1}{2}$$ Second term: $$a_2 = a \times r = 100 \times \frac{1}{2} = 50$$ **Final answers:** - a(i): 1760 mm, 1774 mm, 1788 mm, 1802 mm - a(ii): $a_n = 1760 + (n-1) \times 14$ - a(iii): 1984 mm - b: 24, 12, 6 - c: 50