1. Mia has a geometric sequence of pie slice volumes where the second slice volume is $u_2=30$ cm³ and the fifth slice volume is $u_5=240$ cm³. 2. Recall the general formula for the $n$th term: $$u_n=u_1 r^{n-1}$$ where $r$ is the common ratio and $u_1$ is the smallest slice volume. 3. Using the given terms: $$u_2 = u_1 r = 30$$ and $$u_5 = u_1 r^4 = 240$$. 4. Substitute $u_1 = \frac{30}{r}$ from the first into the second equation: $$\frac{30}{r} r^4 = 240 \implies 30 r^{3} = 240$$. 5. Solve for $r^{3}$: $$r^{3} = \frac{240}{30} = 8$$. 6. Take cube root to find common ratio: $$r = \sqrt[3]{8} = 2$$. (b) Find the smallest slice volume $u_1$: 7. Use $u_2 = u_1 r = 30$ and $r=2$: $$u_1 = \frac{30}{2} = 15 \text{ cm}^3$$. (c) Find the total number of slices from total pie volume 61425 cm³: 8. Use sum of geometric series formula: $$S_n = u_1 \frac{r^{n} - 1}{r - 1} = 61425$$ $$15 \times \frac{2^n -1}{2 - 1} = 61425$$ $$15(2^n -1) = 61425$$ 9. Divide both sides by 15: $$2^n - 1 = \frac{61425}{15} = 4095$$ $$2^n = 4096$$ 10. Since $2^{12} = 4096$, the total number of slices $n=12$. --- 9. Infinite geometric series first term $u_1 = a$, second term: $$u_2 = \frac{1}{4} a^2 - 3a$$ Common ratio is: 11. $$r = \frac{u_2}{u_1} = \frac{\frac{1}{4} a^2 - 3a}{a} = \frac{1}{4}a - 3$$. --- 10.(a) Given sequences: $$u_n: 2, 6, 18, 54, ...$$ common ratio $r_1=3$. $$v_n: 2, -6, 18, -54, ...$$ common ratio $r_2=-3$. 12. The third sequence: $$w_n = u_n + v_n$$ First three non-zero terms: $$w_1 = 2 + 2 = 4$$ $$w_2 = 6 + (-6) = 0$$ $$w_3 = 18 + 18 = 36$$ $$w_4 = 54 + (-54) = 0$$ Next non-zero term is $w_5 = u_5 + v_5 = 162 + 162 = 324$. (b.i) Since non-zero terms occur at odd $n$, pattern is $w_1=4$, $w_3=36$, $w_5=324$. The sequence of these terms has ratio: $$r = \frac{36}{4} = 9$$. (b.ii) Sum rewritten as: $$\sum_{k=1}^{225} w_k = \sum_{k=0}^m 4 r^k = \sum_{k=0}^m 4 \times 9^k$$ Number of terms in $w_n$ non-zero is about half of 225: So $m=112$. --- 11.(a.i) Amelia invests 9000 at 7% compounded annually 5 years: $$A = P(1 + r)^t = 9000(1 + 0.07)^5 = 9000(1.40255) = 12622.95$$ Rounded to nearest hundred: 12600. (a.ii) Years to reach 20000: $$20000 = 9000(1.07)^t$$ $$\frac{20000}{9000} = (1.07)^t$$ $$2.2222 = (1.07)^t$$ Take log: $$t = \frac{\log 2.2222}{\log 1.07} \approx \frac{0.3468}{0.0296} = 11.72$$ Round up: 12 years. (b) Bill invests 9000 at rate $r$ Compounded monthly for 10 years to reach 20000: $$20000 = 9000 \left(1 + \frac{r}{12}\right)^{120}$$ $$\frac{20000}{9000} = \left(1 + \frac{r}{12}\right)^{120} = 2.2222$$ Take logarithm: $$\log 2.2222 = 120 \log \left(1 + \frac{r}{12}\right)$$ Approximate: $$\log(1 + \frac{r}{12}) = \frac{0.3468}{120} = 0.00289$$ $$1 + \frac{r}{12} = 10^{0.00289} \approx 1.00667$$ $$\frac{r}{12} = 0.00667$$ $$r = 0.00667 \times 12 = 0.08004 = 8.00\%$$ (c.i) Chris adds half amount deposited previously each year starting from 9000: Sum after infinite years: $$S = 9000 + 4500 + 2250 + \cdots = \frac{9000}{1 - \frac{1}{2}} = 18000 < 20000$$ Chris never reaches 20000. (c.ii) Find initial deposit $x$ for Chris to reach 20000 after 5 years: $$S_5 = x + \frac{x}{2} + \frac{x}{4} + \frac{x}{8} + \frac{x}{16} = x\left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}\right) = x \times \frac{31}{16}$$ Set sum to 20000: $$x \times \frac{31}{16} = 20000$$ $$x = \frac{20000 \times 16}{31} \approx 10322$$ Final answers: (a) Common ratio $r=2$. (b) Smallest slice volume $u_1=15$ cm³. (c) Total slices $n=12$. 9. Common ratio in terms of $a$: $r = \frac{1}{4}a - 3$. 10.(a) First three non-zero terms of $w_n$: 4, 36, 324. 10.(b.i) $r=9$. 10.(b.ii) $m=112$. 11.(a.i) Amelia’s 5-year amount: 12600. 11.(a.ii) Years to target: 12. 11.(b) Bill’s minimum rate: 8.00%. 11.(c.i) Chris never reaches 20000 with initial 9000. 11.(c.ii) Required initial deposit for Chris: 10322.