1. Calculate the value of $r$ given $r = 2a - \frac{\sqrt{b}}{c}$, $a = 0.975$, $b = 4.41$, $c = 35$. Step 1: Calculate $\sqrt{b} = \sqrt{4.41}$. $$\sqrt{4.41} = 2.1$$ Step 2: Calculate $\frac{\sqrt{b}}{c} = \frac{2.1}{35} = 0.06$. Step 3: Calculate $2a = 2 \times 0.975 = 1.95$. Step 4: Calculate $r = 1.95 - 0.06 = 1.89$. Answer: $r = 1.89$ --- 2.1 Write down $a$, $b$ and $c$ each correct to one significant figure. $a = 0.975 \approx 1$ $b = 4.41 \approx 4$ $c = 35 \approx 40$ 2.2 Find Albert's estimate of the value of $r$ using these one significant figure values. Calculate $\sqrt{b} = \sqrt{4} = 2$ Calculate $\frac{\sqrt{b}}{c} = \frac{2}{40} = 0.05$ Calculate $2a = 2 \times 1 = 2$ Calculate $r = 2 - 0.05 = 1.95$ Answer: Estimate $r = 1.95$ --- 3. Calculate the percentage error in Albert's estimate. Percentage error = $\frac{|\text{True value} - \text{Estimate}|}{\text{True value}} \times 100$ = $\frac{|1.89 - 1.95|}{1.89} \times 100 = \frac{0.06}{1.89} \times 100 \approx 3.17 \%$ Answer: Percentage error is approximately $3.17 \%$ --- 4. Jeremy invests 8000 at 5.5% compounded annually. 4.1 Write formula for value after $n$ years. $$A = 8000(1 + 0.055)^n$$ 4.2 Calculate amount after 1 year and 3 years. After 1 year: $$A = 8000(1.055)^1 = 8000 \times 1.055 = 8440$$ After 3 years: $$A = 8000(1.055)^3 = 8000 \times 1.1746 = 9396.8$$ 4.3 Can Jeremy put down $10 000 deposit in 5 years? Calculate value after 5 years: $$A = 8000(1.055)^5 = 8000 \times 1.307 = 10456$$ Since $10456 > 10000$, Jeremy can afford the deposit within 5 years. Answer: Yes, Jeremy will have enough money. --- 5. Identify arithmetic and geometric sequences: a) $a_n = 1,5,10,15,\ldots$ Differences: 5-1=4,10-5=5 (not constant) - Not arithmetic. b) $b_n = \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5},\ldots$ Differences: $\frac{2}{3} - \frac{1}{2} = \frac{1}{6}$, next difference $\frac{3}{4} - \frac{2}{3} = \frac{1}{12}$ (not constant) - Not arithmetic. c) $c_n = 1.5, 3, 4.5, 6, \ldots$ Differences: 3-1.5 =1.5, 4.5-3=1.5 (constant) - Arithmetic with common difference $d=1.5$. d) $d_n = 2, 1, \frac{1}{2}, \frac{1}{4}, \ldots$ Ratios: $\frac{1}{2} = \frac{1}{2} \times 2$, $\frac{1}{2}\div 1 = \frac{1}{2}$, $\frac{1}{4} \div \frac{1}{2} = \frac{1}{2}$ (constant) - Geometric with common ratio $r=\frac{1}{2}$. 5.1 Arithmetic sequence is $c_n$ with $d=1.5$. 5.2 Geometric sequence is $d_n$ with $r=\frac{1}{2}$. 5.3 Find $8^{th}$ term of geometric sequence $d_n$. General term: $u_n = u_1 \times r^{n-1} = 2 \times (\frac{1}{2})^{7} = 2 \times \frac{1}{128} = \frac{1}{64}$. Answer: $u_8 = \frac{1}{64}$. --- 6. (For Question 14 sequences) Arithmetic: $b_n = 2.5, 5, 7.5, 10, ...$ difference $d=2.5$. Geometric: $c_n = 3, 1, \frac{1}{3}, \frac{1}{9}, ...$ ratio $r=\frac{1}{3}$. Find $6^{th}$ term of geometric sequence: $$u_6 = u_1 \times r^{5} = 3 \times \left(\frac{1}{3}\right)^5 = 3 \times \frac{1}{243} = \frac{1}{81}$$ --- 7. Arithmetic sequence with $u_1=40, u_2=32, u_3=24$. 7.1 Common difference: $d = u_2 - u_1 = 32 - 40 = -8$ 7.2 Find $u_8$: $$u_8 = u_1 + 7d = 40 + 7 \times (-8) = 40 - 56 = -16$$ 7.3 Find sum $S_8$ of first 8 terms: $$S_8 = \frac{8}{2}(u_1 + u_8) = 4 \times (40 -16) = 4 \times 24 = 96$$ --- 8. Arithmetic sequence with $u_1=12, u_2=21, u_3=30$. 8.1 $d=21-12=9$ 8.2 Find $u_{10}$: $$u_{10} = 12 + 9 \times (10-1) = 12 + 81 = 93$$ 8.3 Find sum $S_{10}$: $$S_{10} = \frac{10}{2}(12 + 93) = 5 \times 105 = 525$$ --- 9. Arithmetic sequence with $u_{15} = 21$, $d=-4$. 9.1 Find $u_1$: $$u_{15} = u_1 + 14d \Rightarrow u_1 = u_{15} - 14d = 21 - 14 \times (-4) = 21 + 56 = 77$$ 9.2 Find $u_{29}$: $$u_{29} = u_1 + 28d = 77 + 28 \times (-4) = 77 - 112 = -35$$ 9.3 Sum first 40 terms $S_{40}$: $$S_{40} = \frac{40}{2}(u_1 + u_{40})$$ Find $u_{40}$: $$u_{40} = u_1 + 39d = 77 + 39 \times (-4) = 77 - 156 = -79$$ Sum: $$S_{40} = 20 \times (77 - 79) = 20 \times (-2) = -40$$ --- 10. Geometric sequence with $u_1=5$, $u_2=-1$, $u_3=\frac{1}{5}$. 10.1 Find common ratio $r$: $$r = \frac{u_2}{u_1} = \frac{-1}{5} = -\frac{1}{5}$$ Check: $$u_3 = u_2 \times r = -1 \times (-\frac{1}{5}) = \frac{1}{5}$$ 10.2 Find $u_7$: $$u_7 = u_1 \times r^{6} = 5 \times \left(-\frac{1}{5}\right)^6 = 5 \times \frac{1}{15625} = \frac{5}{15625} = \frac{1}{3125}$$ 10.3 Find sum $S_7$: Formula for sum geometric series: $$S_n = u_1 \times \frac{1-r^n}{1-r}$$ $$S_7 = 5 \times \frac{1 - (-\frac{1}{5})^7}{1 - (-\frac{1}{5})} = 5 \times \frac{1 - (-\frac{1}{78125})}{1 + \frac{1}{5}} = 5 \times \frac{1 + \frac{1}{78125}}{\frac{6}{5}} = 5 \times \frac{\frac{78125 + 1}{78125}}{\frac{6}{5}}$$ $$= 5 \times \frac{78126}{78125} \times \frac{5}{6} = \frac{25 \times 78126}{6 \times 78125} = \frac{25 \times 78126}{468750} \approx 4.17$$ Exact fraction can remain as above. --- 11. Mia deposits 4000 AUD at 6% nominal annual interest compounded semi-annually for 2.5 years. 11.1 Interest earned: Number of periods: $$n = 2.5 \times 2 = 5$$ Rate per period: $$r = \frac{6\%}{2} = 3\% = 0.03$$ Amount: $$A = 4000(1 + 0.03)^5 = 4000 \times 1.159274 = 4637.10$$ Interest earned: $$4637.10 - 4000 = 637.10$$ Answer: $637.10$ AUD interest. --- 12. Ella deposits an amount $P$, at 4% nominal annual interest compounded monthly, amount after 2.5 years is 4000 AUD. Number of monthly periods: $$n = 2.5 \times 12 = 30$$ Monthly interest rate: $$r = \frac{4\%}{12} = \frac{0.04}{12} = 0.0033333$$ Formula: $$4000 = P(1 + 0.0033333)^{30}$$ Calculate $(1 + 0.0033333)^{30}$: $$\approx 1.104941$$ Calculate $P$: $$P = \frac{4000}{1.104941} = 3619.72$$ Answer: Ella deposited approximately 3619.72 AUD. --- 13. Tennis ball bounces height forming geometric sequence: $h_1=80$cm, $h_2=60$cm. 13.1 Find common ratio $r$: $$r = \frac{h_2}{h_1} = \frac{60}{80} = 0.75$$ 13.2 Find height at 10th bounce: $$h_{10} = h_1 \times r^{9} = 80 \times 0.75^9$$ Calculate $0.75^9 \approx 0.0751$ $$h_{10} = 80 \times 0.0751 = 6.01 \text{ cm}$$ 13.3 Total distance for first 6 bounces (up and down): Distance traveled is: Initial drop $h_1 = 80$ cm Then up and down 5 times: $$2 \times (h_2 + h_3 + h_4 + h_5 + h_6)$$ Calculate sum: $$S = h_2 + h_3 + h_4 + h_5 + h_6 = h_1 r + h_1 r^2 + h_1 r^3 + h_1 r^4 + h_1 r^5$$ $$= 80 (r + r^2 + r^3 + r^4 + r^5)$$ Sum of geometric series: $$\sum_{k=1}^5 r^k = r \times \frac{1 - r^5}{1 - r} = 0.75 \times \frac{1 - 0.2373}{1 - 0.75} = 0.75 \times \frac{0.7627}{0.25} = 0.75 \times 3.0508 = 2.288$$ Calculate total distance: $$80 + 2 \times 80 \times 2.288 = 80 + 2 \times 183.04 = 80 + 366.08 = 446.08 \text{ cm}$$ Answer: Total distance is $446.08$ cm.