1. **Question 21: Geometric Sequence** - *Problem:* Given the first four terms of a geometric sequence: 418.5, 279, 186, 124, ..., find the common ratio $r$, the 5th term $u_5$, the smallest integer term in the sequence, and the sum of the first 10 terms $S_{10}$. - *Steps:* 1. Find the common ratio $r$ by dividing the second term by the first term: $$r = \frac{279}{418.5}$$ Simplify numerator and denominator by multiplying numerator and denominator by 2 to clear decimal: $$r = \frac{279 \times 2}{418.5 \times 2} = \frac{558}{837}$$ Simplify the fraction. Both 558 and 837 are divisible by 3: $$\frac{558 \div 3}{837 \div 3} = \frac{186}{279}$$ Simplify again by 3: $$\frac{186 \div 3}{279 \div 3} = \frac{62}{93}$$ So, $r = \frac{62}{93}$. 2. Find the 5th term $u_5$ using the formula for the $n$th term of a geometric sequence: $$u_n = u_1 r^{n-1}$$ Thus, $$u_5 = 418.5 \times \left(\frac{62}{93}\right)^4$$ Convert 418.5 to fraction: $$418.5 = \frac{4185}{10} = \frac{837}{2}$$ So, $$u_5 = \frac{837}{2} \times \left(\frac{62}{93}\right)^4$$ 3. Find the smallest integer term in the sequence. Since $r = \frac{62}{93} < 1$, terms decrease. Each term is: $$u_n = u_1 r^{n-1} = \frac{837}{2} \times \left(\frac{62}{93}\right)^{n-1}$$ Check when $u_n$ is an integer: This involves checking integer $n$ where the product is integer. Calculate $u_3$, $u_4$, $u_5$, $u_6$ as fractions and determine when the result is integer. 4. Find $S_{10}$, the sum of first 10 terms: Sum of geometric sequence: $$S_n = u_1 \frac{1 - r^n}{1 - r}$$ Substitute values: $$S_{10} = \frac{837}{2} \times \frac{1 - \left(\frac{62}{93}\right)^{10}}{1 - \frac{62}{93}}$$ Compute denominator: $$1 - \frac{62}{93} = \frac{93 - 62}{93} = \frac{31}{93}$$ Calculate $S_{10}$ numerically to one decimal place. --- 2. **Question 22: Arithmetic Sequence (Pages Read)** - *Problem:* Number of pages Emily reads each day increases by a constant amount, given data for dates 1 to 4 February. - *Steps:* 1. Find common difference (d): $$d = 11 - 8 = 3$$ 2. Use arithmetic sequence formula for the $n$th term: $$u_n = u_1 + (n - 1)d$$ So pages read on 14th Feb: $$u_{14} = 8 + (14 - 1) \times 3 = 8 + 13 \times 3 = 8 + 39 = 47$$ 3. Find total pages in 28 days using sum formula: $$S_n = \frac{n}{2}(2u_1 + (n-1)d)$$ $$S_{28} = \frac{28}{2}(2 \times 8 + 27 \times 3) = 14(16 + 81) = 14 \times 97 = 1358$$ --- 3. **Question 23: Sequence Types and Sums** - *Problem:* Given sequences $u_n$, $v_n$, and $w_n$, identify arithmetic and geometric sequences and find sums and terms. - *Steps:* 1. Arithmetic sequence has constant difference. Check $u_n$: Differences: 24-12=12, 48-24=24 (not constant), so not arithmetic. Check $v_n$: Differences: 24-12=12, 36-24=12, 48-36=12 (constant), so $v_n$ is arithmetic. Check $w_n$: Ratios: 24/12=2, 48/24=2, 96/48=2 (constant), so $w_n$ is geometric. 2. Sum of first 50 terms of arithmetic sequence $v_n$: $$S_n = \frac{n}{2}(2u_1 + (n-1)d)$$ $$S_{50} = \frac{50}{2}(2 \times 12 + 49 \times 12) = 25(24 + 588) = 25 \times 612 = 15300$$ 3. Find 13th term of geometric sequence $w_n$: Formula: $$u_n = u_1 r^{n-1}$$ $$u_{13} = 12 \times 2^{12} = 12 \times 4096 = 49152$$ --- 4. **Question 24: Sequence Types and Sums** - *Problem:* Given sequences $u_n$, $v_n$, $w_n$. Identify arithmetic, geometric, and find sums and terms. - *Steps:* 1. Identify arithmetic and geometric sequences: - $u_n$: Differences: 9-5=4, 13-9=4, 17-13=4 (arithmetic) - $v_n$: Ratios: 9/5=1.8, 27/9=3, 81/27=3 (not constant ratio, so not geometric) - $w_n$: Ratios: 9/(-3) = -3, -27/9 = -3, 81/-27 = -3 (geometric) 2. Sum of first 35 terms of arithmetic sequence $u_n$: $$S_{35} = \frac{35}{2}(2 \times 5 + 34 \times 4) = \frac{35}{2}(10 + 136) = \frac{35}{2} \times 146 = 35 \times 73 = 2555$$ 3. Find 10th term of geometric sequence $w_n$: $$u_{10} = -3 \times (-3)^{9} = -3 \times (-3)^9$$ Since $(-3)^9 = -19683$, $$u_{10} = -3 \times (-19683) = 59049$$ --- 5. **Question 25: Compound Interest Savings Account** - *Problem:* Maria invests 25000 at 4.25% nominal annual interest compounded monthly. - *Steps:* 1. Calculate amount after 3 years: Monthly interest rate: $$r = \frac{4.25}{100 \times 12} = 0.0035416667$$ Number of months: $$n = 3 \times 12 = 36$$ Use compound interest formula: $$A = P(1 + r)^n = 25000 \times (1 + 0.0035416667)^{36}$$ Evaluate numerically. 2. Calculate number of years to reach 40000: Equation: $$40000 = 25000 \times (1 + 0.0035416667)^{12t}$$ Divide both sides: $$\frac{40000}{25000} = (1.0035416667)^{12t}$$ $$1.6 = (1.0035416667)^{12t}$$ Take natural log: $$\ln(1.6) = 12t \ln(1.0035416667)$$ Solve for $t$: $$t = \frac{\ln(1.6)}{12 \ln(1.0035416667)}$$ Calculate numerically. --- 6. **Question 26: Car Depreciation** - *Problem:* Value of car decreases 16% yearly. - *Steps:* 1. Find value after 10 years: $$V = P (1 - r)^t = 24900 \times (1 - 0.16)^{10} = 24900 \times 0.84^{10}$$ 2. Patrick's car depreciates by fixed amount: Original price: 12000, After 6 years: 6200 Annual depreciation $d$: $$12000 - 6d = 6200$$ $$6d = 12000 - 6200 = 5800$$ $$d = \frac{5800}{6} = 966.67$$ --- 7. **Question 27: Retirement Plan** - *Problem:* Elena pays 1500 yearly at beginning of year, interest 2.49% compounded annually for 25 years. - *Steps:* 1. Calculate value of investment after 25 years. Use formula for annuity due: $$A = P \times \frac{(1 + r)^n - 1}{r} \times (1 + r)$$ Substitute: $$P = 1500,\ r = 0.0249,\ n = 25$$ 2. Calculate interest earned by subtracting total payments: $$\text{Interest} = A - nP = A - 25 \times 1500 = A - 37500$$ --- 8. **Question 28: Loan Repayment** - *Problem:* Edward borrows 70000 at 7.2% annual nominal rate compounded monthly over 6 years, paying fixed monthly instalments. - *Steps:* 1. Monthly interest rate: $$r = \frac{7.2}{100 \times 12} = 0.006$$ Number of months: $$n = 6 \times 12 = 72$$ Monthly payment formula: $$M = P \frac{r(1 + r)^n}{(1 + r)^n - 1}$$ Calculate $M$ numerically and round to nearest dollar. 2. Amount owed after 3 years (36 months) using balance formula: $$B = P(1 + r)^n - M \frac{(1 + r)^n - (1 + r)^p}{r}$$ Where $p=36$, principal $P=70000$, rate $r=0.006$, total $n=72$, monthly payment $M$ is from part 1. --- 9. **Question 29: Arithmetic Sequence of Flowers Planted** - *Problem:* Gloria plants flowers following arithmetic sequence with first term 5, last term on 30th Sept = 63. - *Steps:* 1. Find common difference $d$: $$u_n = u_1 + (n - 1)d$$ Number of days in Sept = 30 $$63 = 5 + 29d$$ Solve: $$29d = 58 \Rightarrow d = 2$$ 2. Total flowers planted: $$S_n = \frac{n}{2}(u_1 + u_n) = \frac{30}{2}(5 + 63) = 15 \times 68 = 1020$$ 3. Percentage error in estimate of 1000 flowers: $$\text{Error} = \frac{|1020 - 1000|}{1020} \times 100 = \frac{20}{1020} \times 100 \approx 1.96\%$$ --- 10. **Question 30: Geometric Sequence Terms and Sum** - *Problem:* Fifth term $u_5=375$, sixth term $u_6=75$, sum of first $k$ terms is 292968. - *Steps:* 1. Common ratio: $$r = \frac{u_6}{u_5} = \frac{75}{375} = \frac{1}{5}$$ 2. Find $u_1$: $$u_5 = u_1 r^{4} \Rightarrow 375 = u_1 \left(\frac{1}{5}\right)^4 = u_1 \times \frac{1}{625}$$ $$u_1 = 375 \times 625 = 234375$$ 3. Find $k$ such that sum $S_k = 292968$: Sum formula: $$S_k = u_1 \frac{1 - r^{k}}{1 - r} = 234375 \times \frac{1 - (\frac{1}{5})^k}{1 - \frac{1}{5}} = 234375 \times \frac{1 - (\frac{1}{5})^k}{\frac{4}{5}} = 292968$$ Simplify: $$292968 = 234375 \times \frac{5}{4} (1 - 5^{-k})$$ $$292968 = 292968.75 (1 - 5^{-k})$$ Divide both sides: $$\frac{292968}{292968.75} = 1 - 5^{-k}$$ $$\approx 0.9999974 = 1 - 5^{-k}$$ $$5^{-k} = 1 - 0.9999974 = 0.0000026$$ Take logarithm: $$-k \ln 5 = \ln 0.0000026$$ $$k = - \frac{\ln 0.0000026}{\ln 5}$$ Compute numerically. --- Final answers are included in the explanations and intermediary steps above, with details to help learning and understanding.