1. **Problem statement:** (a)(i) The first term $a_1$ of an arithmetic progression (AP) is 3 and the sum of its 8 terms $S_8$ is 164. Find the common difference $d$. (a)(ii) Given the sum of the first $n$ terms $S_n$ is 570, find $n$. (b)(i) The first, fifth, and seventh terms of another AP form a decreasing geometric progression (GP) with the first term of the GP as 64. Find the common difference of this AP. (b)(ii) Find the sum of the first 10 terms of the GP. --- 2. **Formulas and rules:** - Sum of first $n$ terms of an AP: $$S_n = \frac{n}{2} [2a_1 + (n-1)d]$$ - $n$th term of AP: $$a_n = a_1 + (n-1)d$$ - Terms of GP satisfy: $$t_2^2 = t_1 \times t_3$$ - Sum of first $n$ terms of GP: $$S_n = a \frac{1-r^n}{1-r}$$ where $a$ is first term and $r$ common ratio. --- 3. **Solution:** (a)(i) Given $a_1=3$, $S_8=164$, find $d$. Use sum formula: $$164 = \frac{8}{2} [2(3) + (8-1)d] = 4[6 + 7d] = 24 + 28d$$ Solve for $d$: $$164 = 24 + 28d \implies 28d = 140 \implies d = 5$$ (a)(ii) Given $S_n=570$, find $n$ with $a_1=3$, $d=5$: Sum formula: $$570 = \frac{n}{2} [2(3) + (n-1)5] = \frac{n}{2} [6 + 5n - 5] = \frac{n}{2} (5n + 1)$$ Multiply both sides by 2: $$1140 = n(5n + 1) = 5n^2 + n$$ Rearranged quadratic: $$5n^2 + n - 1140 = 0$$ Use quadratic formula: $$n = \frac{-1 \pm \sqrt{1 + 4 \times 5 \times 1140}}{2 \times 5} = \frac{-1 \pm \sqrt{1 + 22800}}{10} = \frac{-1 \pm \sqrt{22801}}{10}$$ Since $n$ must be positive integer, approximate: $$\sqrt{22801} \approx 151$$ So: $$n = \frac{-1 + 151}{10} = \frac{150}{10} = 15$$ (b)(i) Let the AP have first term $A$ and common difference $d$. The terms are: $$a_1 = A$$ $$a_5 = A + 4d$$ $$a_7 = A + 6d$$ These form a decreasing GP with first term 64, so: $$a_1 = 64$$ GP condition: $$a_5^2 = a_1 \times a_7$$ Substitute: $$(A + 4d)^2 = A (A + 6d)$$ Plug $A=64$: $$(64 + 4d)^2 = 64 (64 + 6d)$$ Expand left: $$4096 + 512d + 16d^2 = 4096 + 384d$$ Bring all terms to one side: $$16d^2 + 512d - 384d = 0 \implies 16d^2 + 128d = 0$$ Divide by 16: $$d^2 + 8d = 0 \implies d(d + 8) = 0$$ So: $$d = 0 \text{ or } d = -8$$ Since the GP is decreasing, common difference $d = -8$. (b)(ii) The GP terms are: $$t_1 = 64$$ $$t_2 = a_5 = 64 + 4(-8) = 64 - 32 = 32$$ Common ratio: $$r = \frac{t_2}{t_1} = \frac{32}{64} = \frac{1}{2}$$ Sum of first 10 terms of GP: $$S_{10} = 64 \times \frac{1 - (\frac{1}{2})^{10}}{1 - \frac{1}{2}} = 64 \times \frac{1 - \frac{1}{1024}}{\frac{1}{2}} = 64 \times 2 \times \left(1 - \frac{1}{1024}\right) = 128 \times \frac{1023}{1024} = \frac{128 \times 1023}{1024}$$ Simplify: $$S_{10} = 128 \times \left(1 - \frac{1}{1024}\right) = 128 - \frac{128}{1024} = 128 - \frac{1}{8} = 127.875$$ --- **Final answers:** (a)(i) $d = 5$ (a)(ii) $n = 15$ (b)(i) $d = -8$ (b)(ii) $S_{10} = 127.875$