4.7 Sequences and series in modelling — Practise with help

Work each one first, closed book. Decide whether the quantity changes by a fixed amount or by a fixed factor before you write anything down. State the assumptions of your model in one sentence.

Warm-up: remember

Work out the 5th term of the geometric sequence $3, 6, 12, \dots$ .
Work out the sum of the first 6 terms of $2, 5, 8, \dots$ .
A recurrence is given by $u_{n+1} = 0.5u_n + 10$ and $u_1 = 10$ . Work out $u_3$ .

Answers

Here $a = 3$ and $r = 2$ , so $u_5 = 3(2)^4 = 48$ .
The sequence is arithmetic with $a = 2$ and $d = 3$ , so $S_6 = \frac{6}{2}(2(2) + 5(3)) = 3(19) = 57$ .
$u_2 = 0.5(10) + 10 = 15$ , so $u_3 = 0.5(15) + 10 = 17.5$ .

If question 3 was hard, revisit recurrences in Section 4.3 before you continue. The recurrence adds a fixed amount each step, so the terms settle.

Practise with help

1. A quantity takes the values $200, 220, 242, \dots$ . Decide whether an arithmetic or a geometric model fits, and work out the 6th value.

Hint

Work out the differences and the ratios between consecutive values. One of them is constant.

Next step

The ratios are $\frac{220}{200} = 1.1$ and $\frac{242}{220} = 1.1$ , so the model is geometric with $r = 1.1$ . Then $u_6 = 200(1.1)^5$ .

Full working

The ratios are constant, so the model is geometric with $a = 200$ and $r = 1.1$ .

u_6 = 200(1.1)^5 = 200(1.61051) = 322.102.

The 6th value is $322.102$ . The differences grow, which is the sign of a geometric model: $20$ , then $22$ , then more each step.

2. A quantity takes the values $80, 95, 110, \dots$ . Decide which model fits, and work out the 10th value.

Hint

Check the differences first. If they are constant, the model is arithmetic.

Next step

The difference is $95 - 80 = 15$ , and $110 - 95 = 15$ . Use $u_n = 80 + (n - 1)(15)$ with $n = 10$ .

Full working

The differences are constant at $15$ , so the model is arithmetic with $a = 80$ and $d = 15$ .

u_{10} = 80 + (10 - 1)(15) = 80 + 135 = 215.

The 10th value is $215$ . The model adds the same amount at every step, so a graph of the values is a straight line.

3. At the start of year 1 a village has $12{,}000$ people, and the population grows by 3% each year. Work out the population at the start of year 6.

Hint

A rise of 3% means the growth factor is $1.03$ . This is a geometric model, so use $P_n = P_1 r^{n-1}$ .

Next step

$P_6 = 12{,}000(1.03)^{6-1} = 12{,}000(1.03)^5$ . Work out $1.03^5$ first, then multiply.

Full working

P_6 = 12{,}000(1.03)^5 = 12{,}000(1.1592740743) = 13{,}911.3.

The population is about $13{,}911$ people at the start of year 6. A population is a whole number, so the value is rounded, and the model treats a fraction of a person as meaningful, which it cannot be.

4. A tree is $2$ m tall at the start of year 1 and grows by 8% each year, so $h_n = 2(1.08)^{n-1}$ metres at the start of year $n$ . Use trial and improvement to find when the tree first passes $3$ m.

Hint

You need $2(1.08)^{n-1} > 3$ , so $(1.08)^{n-1} > 1.5$ . Try two values of the power on either side of $1.5$ .

Next step

Work out $1.08^5$ and $1.08^6$ , then substitute each into the model to find the two heights around $3$ m.

Full working

1.08^5 = 1.4693, \qquad 1.08^6 = 1.5869.

h_6 = 2(1.4693) = 2.939 \text{ m}, \qquad h_7 = 2(1.5869) = 3.174 \text{ m}.

The tree is below $3$ m at the start of year 6 and above it at the start of year 7, so it first passes $3$ m during year 6. The answer names a year, and the model gives no information about the height at a finer interval.

5. A patient takes a dose each day. The amount in the body just after a dose follows $u_{n+1} = 0.4u_n + 60$ with $u_1 = 60$ mg. Find the level the amount settles on.

Hint

At the settling level $L$ , the next amount equals the current one, so $L = 0.4L + 60$ .

Next step

Subtract $0.4L$ from both sides: $0.6L = 60$ . Then divide by $0.6$ .

Full working

L = 0.4L + 60 \implies 0.6L = 60 \implies L = 100.

The amount settles at 100 mg. The terms $60, 84, 93.6, \dots$ creep towards $100$ , which the recurrence confirms in one line.

6. A savings account holds £500. Each year the account adds 5% interest, then the saver deposits £200. 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.

Hint

The 4th deposit is three steps after the 1st. Apply the rule three times, or write the terms out one at a time.

Next step

$u_2 = 1.05(500) + 200 = 725$ , then find $u_3$ , then $u_4$ . Keep the pence at each step rather than rounding.

Full working

u_2 = 1.05(500) + 200 = 725,

u_3 = 1.05(725) + 200 = 961.25,

u_4 = 1.05(961.25) + 200 = 1209.3125.

The balance is £1209.31 just after the 4th deposit, to the nearest penny. The interest is worked out before the deposit is added, which is the order the situation describes.

7. The quiz night raises $25n^2 + 125n$ pounds in its first $n$ months. The appeal raises $3000(1 - 0.8^n)$ pounds. Work out the total after 14 months, to the nearest penny.

Hint

Work out each stream for $n = 14$ and add them. The quiz total grows quickly, because it contains $n^2$ .

Next step

Quiz: $25(14^2) + 125(14) = 4900 + 1750$ . Appeal: $3000(1 - 0.8^{14})$ , and $0.8^{14} = 0.0439804651$ to ten decimal places.

Full working

\text{quiz} = 25(196) + 1750 = 6650,

\text{appeal} = 3000(1 - 0.0439804651) = 3000(0.9560195349) = 2868.06.

The total is

6650 + 2868.06 = 9518.06,

which is £9518.06 to the nearest penny. That is above £9,000, so the roof is paid for by the end of month 14.