Sequences and Series in Modelling

Applying sequences and series to real-world financial and growth problems

SPEC 4.6 • SEQUENCES AND SERIES

Choosing Your Model

Many real-world situations involve quantities that change repeatedly over time. The key is recognising which type of sequence to use:

ARITHMETIC

Use when the quantity changes by a fixed amount each period

Next = Previous + constant

Examples: Regular fixed savings, linear salary increases, straight-line depreciation

GEOMETRIC

Use when the quantity changes by a fixed percentage each period

Next = Previous × constant

Examples: Compound interest, percentage depreciation, population growth

Exam Tip

Look for key words: "fixed amount" or "increases by £X" suggests arithmetic; "percentage" or "multiplied by" suggests geometric.

Arithmetic Models

When someone saves or pays a fixed amount each period (with no compound interest), the total forms an arithmetic series.

Regular Fixed Savings

If you deposit £

d

each month for

n

months:

S_{n} = n d

More generally, if amounts increase by a fixed amount each period (first deposit

a

, common difference

d

):

S_{n} = \frac{n}{2} (2 a + (n - 1) d)

or

S_{n} = \frac{n}{2} (a + l)

Key Concept

If each payment is the same, the total is simply

S_{n} = n d

. If payments increase by a fixed amount each time, use the full arithmetic series formula.

01Worked Example

Jasmine starts a savings plan. In the first month she saves £150. Each subsequent month she saves £10 more than the previous month.

(a) Find how much she saves in the 12th month.

(b) Find the total amount saved after 2 years.

1

Identify: Payments increase by fixed £10, so arithmetic with

a = 150

,

d = 10

.

2

(a)

u_{12} = a + (n - 1) d = 150 + 11 (10) = 150 + 110 = £260

3

(b) 2 years = 24 months.

S_{24} = \frac{24}{2} (2 (150) + 23 (10)) = 12 (300 + 230) = £6360

Answers: (a) £260 (b) £6360

Compound Interest

When an investment grows by a fixed percentage each period, the values form a geometric sequence.

Compound Interest Formula

A = P (1 + r)^{n}

Where:

A

= final amount,

P

= principal,

r

= rate per period (decimal),

n

= number of periods

Connection to Geometric Sequences

This is the

n

th term of a geometric sequence with first term

a = P

and common ratio

r = (1 + interest rate)

Important

Make sure

r

and

n

use the same time period! If interest is 6% per annum compounded monthly, then

r = 0.06/12 = 0.005

per month.

02Worked Example

£5000 is invested in an account paying 4% compound interest per annum.

(a) Find the value of the investment after 8 years.

(b) Find the total interest earned.

1

Identify:

P = 5000

,

r = 0.04

,

n = 8

2

(a)

A = 5000 (1.04)^{8} = 5000 \times 1.3686... = £6842.85

3

(b) Interest =

6842.85 - 5000 = £1842.85

Answers: (a) £6842.85 (b) £1842.85

Depreciation

When an asset loses a fixed percentage of its value each period, this is modelled by a geometric sequence.

Depreciation Formula

V = V_{0} (1 - r)^{n}

Where:

V

= value after

n

periods,

V_{0}

= initial value,

r

= depreciation rate (decimal)

Key Difference

For growth: common ratio =

(1 + r) > 1

For decay/depreciation: common ratio =

(1 - r) < 1

Common Applications

Car values, machinery, equipment, radioactive decay, drug levels in bloodstream

03Worked Example

A car is bought for £24,000. It depreciates by 18% each year.

(a) Find the value of the car after 5 years.

(b) After how many complete years will the car first be worth less than £8000?

1

V_{0} = 24000

, depreciation = 18%, so common ratio =

0.82

2

(a)

V = 24000 (0.82)^{5} = 24000 \times 0.3707... = £8898.34

3

(b) Need

24000 (0.82)^{n} < 8000

, so

(0.82)^{n} < \frac{1}{3}

Taking logs:

n lo g (0.82) < lo g (\frac{1}{3})

Since

lo g (0.82) < 0

, inequality reverses:

n > \frac{l o g ( 1/3 )}{l o g ( 0.82 )} = 5.53...

4

Since

n > 5.53

and must be whole years,

n = 6

Answers: (a) £8898.34 (b) 6 years

Other Applications

Geometric sequences model many real-world phenomena beyond finance:

ApplicationModelCommon Ratio

Population Growth

P = P_{0} (1 + r)^{n}

(1 + r) > 1

Radioactive Decay

M = M_{0} (1 - r)^{n}

(1 - r) < 1

Drug in BloodstreamAmount remaining after each doseDecay factor

< 1

Bouncing BallHeights:

h, h r, h r^{2}, ...

r < 1

Geometric Series

When you need a total (e.g., total distance travelled by bouncing ball), use the sum formula.

04Worked Example

The population of a town is 45,000. It is increasing at 2.5% per year.

(a) Find the population after 15 years.

(b) After how many complete years will the population first exceed 60,000?

1

Geometric growth:

P_{0} = 45000

, common ratio =

1.025

2

(a)

P = 45000 (1.025)^{15} = 45000 \times 1.4483... = 65, 173

3

(b) Need

45000 (1.025)^{n} > 60000

, so

(1.025)^{n} > \frac{4}{3}

n lo g (1.025) > lo g (\frac{4}{3})

n > \frac{l o g ( 4/3 )}{l o g ( 1.025 )} = 11.65...

4

Population first exceeds 60,000 after 12 complete years.

Answers: (a) 65,173 (b) 12 years

Worked Example05

A ball is dropped from 2 metres. After each bounce, it reaches 75% of its previous height. Find the total vertical distance travelled before it comes to rest.

1

Ball travels: 2m down, then up and down for each bounce height.

2

Total =

2 + 2 (2 \times 0.75 + 2 \times 0.7 5^{2} + ...) = 2 + 4 (0.75 + 0.7 5^{2} + ...)

3

The series

0.75 + 0.7 5^{2} + ...

has

a = 0.75

,

r = 0.75

Sum to infinity:

S_{\infty} = \frac{0.75}{1 - 0.75} = \frac{0.75}{0.25} = 3

4

Total distance =

2 + 4 (3) = 14

metres

Answer14 metres

Model Assumptions and Limitations

When using sequences and series to model real situations, be aware of the assumptions:

Common Assumptions

Constant rate: Interest rates, growth rates remain fixed (in reality, they change)

Discrete time periods: Changes at fixed intervals (reality may be continuous)

No external factors: Model ignores taxes, fees, inflation

Infinite continuation: Sum to infinity assumes process continues forever

When Models Break Down

Very long time periods (rates change)

Population models (resources limit growth)

When rate depends on current value

Practice

Question 1: A company offers two bonus options:

Option A: £1000 now, increasing by £100 each year for 8 years

Option B: £800 now, increasing by 12% each year for 8 years

Find the total under each option and state which is better.

Option A (Arithmetic):

S_{8} = \frac{8}{2} (2 (1000) + 7 (100)) = 4 (2700) = £10, 800

Option B (Geometric):

S_{8} = \frac{800 ( 1.1 2 ^{8} - 1 )}{0.12} = \frac{800 ( 1.476... )}{0.12} = £9838.32

AnswerOption A (£10,800) is better by £961.68

Practice

Question 2: Equipment worth £50,000 depreciates by 15% each year. After how many complete years will it first be worth less than 20% of its original value?

Need

50000 (0.85)^{n} < 10000

, so

(0.85)^{n} < 0.2

Taking logs:

n lo g (0.85) < lo g (0.2)

Since

lo g (0.85) < 0

:

n > \frac{l o g ( 0.2 )}{l o g ( 0.85 )} = 9.90...

Answer10 complete years

Practice

Question 3: A bacteria culture doubles every 3 hours. Initially there are 500 bacteria.

(a) How many bacteria after 24 hours?

(b) After how many hours will the population first exceed 1 million?

(a) 24 hours = 8 periods.

P = 500 \times 2^{8} = 128, 000

bacteria

(b) Need

500 \times 2^{n} > 1, 000, 000

, so

2^{n} > 2000

n > \frac{l o g 2000}{l o g 2} = 10.97...

, so

n = 11

periods = 33 hours

Answers: (a) 128,000 (b) 33 hours

Summary

Choosing the Right Model

SituationModelFormula

Fixed amount addedArithmetic

S_{n} = \frac{n}{2} (2 a + (n - 1) d)

Fixed % growthGeometric

A = P (1 + r)^{n}

Fixed % decayGeometric

V = V_{0} (1 - r)^{n}

Total neededGeo. Series

S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}

or

S_{\infty} = \frac{a}{1 - r}

Key Exam Skills

Identify arithmetic vs geometric from context

Use logarithms to find number of terms when needed

Remember inequality reverses when dividing by negative log

Interpret answers in context and consider model limitations