4.7 Sequences and series in modelling — Practise

Work through these on your own. Decide whether the quantity changes by a fixed amount or a fixed factor. Write the model down, say what it assumes, and only then calculate.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

One list is $300, 330, 363, \dots$ and another is $300, 340, 380, \dots$ . For each list, decide whether an arithmetic or a geometric model fits, and give the common difference or common ratio.
A quantity follows the geometric model $u_n = 300(1.1)^{n-1}$ . Work out the 8th value, to 2 decimal places.
A quantity follows the arithmetic model $u_n = 300 + (n - 1)40$ . Work out the 15th value.
A town has $25{,}000$ people at the start of year 1, and the population falls by 4% each year. Work out the population at the start of year 10, to the nearest person.
A recurrence model is $u_{n+1} = 0.7u_n + 90$ . Find the level the terms settle on.
A van costs £18,000. Its value falls by 12% each year. Work out its value after 4 years, to the nearest penny.
A saver starts with £500. Each year the balance grows by 5%, then £200 is deposited. The balance just after a deposit follows $u_{n+1} = 1.05u_n + 200$ with $u_1 = 500$ . Work out the balance just after the 4th deposit, to the nearest penny.
A village has $4000$ people at the start of year 1 and grows by 2% each year, so $P_n = 4000(1.02)^{n-1}$ . Use trial and improvement to find when the population first passes $4500$ .
A fund has two plans. Plan A pays £1000 in month 1 and £60 more each month. Plan B pays £1500 in month 1 and 5% more each month. Work out the total paid by each plan over the first 5 months, and state which is larger.
An investment of £1200 grows by 7% each year. Work out its value after 5 years, to the nearest penny.
A population model is $P_n = 4000(1.02)^{n-1}$ . Measurements suggest that the growth rate was too high, so the model is refined to $P_n = 4000(1.015)^{n-1}$ . Work out $P_7$ under the refined model, to the nearest whole number, and compare it with the original.
The roof fund has a quiz total of $25n^2 + 125n$ pounds and an appeal total of $3000(1 - 0.8^n)$ pounds after $n$ months. Show that the fund passes £9,000 during month 14.

Answers and concise solutions

1. The two models

Answer: For $300, 330, 363, \dots$ the ratios are $\frac{330}{300} = 1.1$ and $\frac{363}{330} = 1.1$ , which are constant, so the model is geometric with $r = 1.1$ . For $300, 340, 380, \dots$ the differences are $40$ and $40$ , so the model is arithmetic with $d = 40$ . Working out the ratios settles the choice when the numbers grow quickly.

2. The value

Answer: Substitute $n = 8$ : $u_8 = 300(1.1)^{7} = 300(1.9487171) = 584.61513$ , which is $584.62$ to 2 decimal places. The growth factor $1.1$ appears seven times, because the first value is the starting point.

3. The value

Answer: Substitute $n = 15$ : $u_{15} = 300 + 14(40) = 300 + 560 = 860$ . The fixed amount $40$ is added fourteen times, one fewer than the position, because the first value is the starting point.

4. The population

Answer: A fall of 4% leaves 96% each year, so the model is $P_n = 25{,}000(0.96)^{n-1}$ . At the start of year 10, $P_{10} = 25{,}000(0.96)^9 = 17{,}313.35$ , which is $17{,}313$ people to the nearest person. The model treats a fraction of a person as meaningful; the answer is rounded because a population is counted in whole people.

5. The settling level

Answer: At the settling level $L$ , the next term equals the current one, so $L = 0.7L + 90$ . Then $0.3L = 90$ and $L = 300$ . The formula $L = \frac{c}{1 - r}$ gives the same value with $r = 0.7$ and $c = 90$ . The level exists because $|r| < 1$ .

6. The value

Answer: A fall of 12% leaves 88%, so the model is $V_n = 18{,}000(0.88)^{n-1}$ . After 4 years, $V_5 = 18{,}000(0.88)^4 = 18{,}000(0.59969536) = 10{,}794.51648$ , which is £10,794.52 to the nearest penny. The factor $0.88$ appears once for each year that has passed.

7. The balance

Answer: Apply the rule three times from $u_1 = 500$ : $u_2 = 1.05(500) + 200 = 725$ , then $u_3 = 1.05(725) + 200 = 961.25$ , then $u_4 = 1.05(961.25) + 200 = 1209.3125$ . The balance is £1209.31 to the nearest penny. The interest is added before the deposit, which is the order the model describes.

8. The year it passes 4500

Answer: Work out the model on both sides of $4500$ . At the start of year 6, $P_6 = 4000(1.02)^5 = 4416.3$ , which is below. At the start of year 7, $P_7 = 4000(1.02)^6 = 4504.6$ , which is above. So the population first passes $4500$ during year 6. Trial and improvement is the honest method here: it shows exactly where the target lies, and it needs no logarithms.

9. The two totals

Answer: Plan A is arithmetic with $a = 1000$ and $d = 60$ , so $S_5 = \frac{5}{2}(2(1000) + 4(60)) = 2.5(2240) = 5600$ . Plan B is geometric with $a = 1500$ and $r = 1.05$ , so $S_5 = \frac{1500(1.05^5 - 1)}{0.05} = 8288.446875$ . Plan B pays more after 5 months: £8288.45 against £5600. The plan B model assumes 5% growth every month, which is 80% a year and cannot continue indefinitely; that assumption is part of what the plan claims.

10. The value

Answer: A rise of 7% gives the growth factor $1 + 0.07 = 1.07$ , so the model is $V_n = 1200(1.07)^{n-1}$ . After 5 years, $V_6 = 1200(1.07)^5 = 1200(1.40255173) = 1683.06207684$ , which is £1683.06 to the nearest penny. The growth factor is the number the old value is multiplied by, not the number added.

11. The refined value

Answer: Under the refined model, $P_7 = 4000(1.015)^6 = 4373.77$ , which is 4374 people to the nearest person. The original model gives $P_7 = 4000(1.02)^6 = 4504.65$ , about 131 people more. Refining a model changes the numbers, not the method: the same geometric structure stays, and only the growth factor is replaced.

12. The two totals either side of the target

Answer: At the end of month 13 the total is $25(13)^2 + 125(13) + 3000(1 - 0.8^{13}) = 5850 + 2835.07 = 8685.07$ , which is below £9,000. At the end of month 14 it is $25(14)^2 + 125(14) + 3000(1 - 0.8^{14}) = 6650 + 2868.06 = 9518.06$ , which is above. The fund therefore passes £9,000 during month 14. Checking both sides of the target is what pins the answer to a single month.