What is 4.6: Sequences and series in modelling?

A constant additive change suggests an arithmetic model, while a constant multiplier or percentage change suggests a geometric model.

How many marks are 4.6 questions worth?

4.6 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has 4.6 appeared in?

We have not yet indexed a verified past-paper appearance for 4.6 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

4.6

Use sequences and series in modelling.

Draft — not yet indexed

Sequences and series in modelling

Worked answers and methods for 4.6 on Edexcel A-level Maths 9MA0.

Explanation

A constant additive change suggests an arithmetic model, while a constant multiplier or percentage change suggests a geometric model.
State what the term number represents and whether the initial value is $u_1$ or $u_0$ before forming a term or sum.
Repeated deposits, withdrawals or other fixed adjustments can be represented by a recurrence and often rewritten using a finite geometric sum.
Interpret results within the context and identify assumptions such as fixed rates or indefinite continuation.
A common error is to use an infinite sum when the process has not converged or has a finite stopping point.

Worked example

A ball is dropped from a height of $12$ m. After each impact it rebounds to $65\%$ of its previous maximum height. Find the total vertical distance travelled before it comes to rest, giving your answer to $3$ significant figures.

1.The initial downward distance is $12$ m.
2.The rebound heights form a geometric series with first term $12(0.65)=7.8$ and ratio $0.65$ .
3.Each rebound height is travelled once upwards and once downwards, so the total is $12+2[7.8/(1-0.65)]=56.5714\ldots$ m.
4.To $3$ significant figures this is $56.6$ m.

Answer: $56.6$ m

Common mistakes

Don't model a fixed percentage change with an additive arithmetic sequence instead of a multiplicative ratio.
Don't count the initial drop twice or omit either the upward or downward part of each rebound.

Exam tip

Build the distance model by separating the first drop from the two-way geometric rebound distances.

Worked practice

Tier 1 · Easy

Mina saves

\pounds 25

in the first month and increases the amount saved by

\pounds 5

each month. Find the total she saves in the first

6

months.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\pounds 225$	2
Notes
The monthly amounts form an arithmetic sequence with $a=25$ , $d=5$ and sixth term $25+5(5)=50$ . Therefore $S_6=6(25+50)/2=225$ .

(2 marks)

Tier 2 · Standard

A theatre has

25

rows of seats. The first row has

16

seats and each successive row has

2

more seats than the preceding row. Find the total number of seats.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$1000$ seats	3
Notes
The row sizes form an arithmetic sequence with $a=16$ , $d=2$ and $n=25$ . The last row has $16+24(2)=64$ seats. Hence the total is $S_{25}=25(16+64)/2=1000$ seats.

(3 marks)

Tier 3 · Hard

An account initially contains

\pounds 5000

. At the end of each year, after interest of

3\%

has been added,

\pounds 400

is withdrawn. The model is

B_0=5000

and

B_{n+1}=1.03B_n-400

. Derive a formula for

B_n

and find the first value of

n

for which the model predicts a negative balance. State one limitation of the model.

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$B_n=5000(1.03)^n-400\dfrac{(1.03)^n-1}{0.03}$ $n=16$ For example, the model assumes the interest rate and withdrawal remain fixed and permits an unrealistic negative balance.	6
Notes
After $n$ years, the initial balance has grown to $5000(1.03)^n$ . The withdrawals accumulate as $400[1+1.03+\cdots+1.03^{n-1}]=400[(1.03)^n-1]/0.03$ , giving the stated formula. Rearranging, $B_n=40000/3-(25000/3)(1.03)^n$ . A negative balance requires $(1.03)^n>1.6$ , so $n>\ln(1.6)/\ln(1.03)=15.90\ldots$ . The first integer is $n=16$ . In reality the account provider would not continue the same process into a negative balance, and the rate may change.

(6 marks)

Tier 1 · Easy

A tank initially contains

500

litres of water. A model assumes that it loses

8\%

of its contents each hour. Give the predicted volume after

3

hours to

3

significant figures.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$389$ litres	2
Notes
The hourly multiplier is $0.92$ . After $3$ hours the model gives $500(0.92)^3=389.344$ litres, which is $389$ litres to $3$ significant figures.

(2 marks)

Tier 2 · Standard

A mosaic is built in numbered frames. Frame

1

uses

18

tiles, and each new frame uses

3

more tiles than the preceding frame. There are

1500

tiles available. Find the greatest number of complete frames that can be built and the number of tiles left unused.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$26$ complete frames $57$ tiles unused	4
Notes
The total for $n$ frames is $S_n=\dfrac n2[2(18)+3(n-1)]=\dfrac{3n(n+11)}{2}$ . Now $S_{26}=3(26)(37)/2=1443$ , while $S_{27}=3(27)(38)/2=1539>1500$ . Therefore $26$ complete frames can be built, leaving $1500-1443=57$ tiles.

(4 marks)

Tier 3 · Hard

Sponsor A contributes

\pounds 200

in the first year and increases its annual contribution by

\pounds 25

each year. Sponsor B contributes

\pounds 400

in the first year and then contributes

90\%

of the previous year's amount in each later year. Find the percentage of the combined total contribution over the first

8

years that comes from Sponsor B. Give your answer to

3

significant figures. Find also the limiting total contributed by Sponsor B.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$49.8\%$ Limiting total from Sponsor B $=\pounds 4000$	6
Notes
Sponsor A contributes $S_8=\dfrac82[2(200)+7(25)]=2300$ . Sponsor B contributes $S_8=400(1-0.9^8)/(1-0.9)=2278.13116$ . Therefore Sponsor B's percentage of the combined total is $100(2278.13116)/(2300+2278.13116)=49.7611\ldots\%$ , which is $49.8\%$ to $3$ significant figures. Since $\|0.9\|<1$ , Sponsor B's limiting total is $400/(1-0.9)=\pounds 4000$ .

(6 marks)

Tier 2 · Standard

The temperature

T_n

degrees Celsius of a machine after

n

ten-minute intervals is modelled by

T_0=84

and

T_{n+1}=0.75T_n+5

. Find

T_3

and determine the steady temperature predicted by the model. State one limitation of the model.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$T_3=47^\circ\mathrm{C}$ Steady temperature $=20^\circ\mathrm{C}$ For example, the model assumes an unchanged surrounding temperature and cooling multiplier.	4
Notes
Iteration gives $T_1=68$ , $T_2=56$ and $T_3=47$ . Since $\|0.75\|<1$ , the recurrence converges to its fixed point. At a steady temperature $T$ , $T=0.75T+5$ , so $0.25T=5$ and $T=20^\circ\mathrm{C}$ . The fixed recurrence ignores changes in the surroundings or in the rate of cooling.

(4 marks)

Tier 3 · Hard

Stage

1

of a cable display has

6

segments, each

9

metres long. To create the next stage, every segment is replaced by two segments, each one third as long. The displays for different stages are built separately. Find the number of segments and the total cable length in stage

n

. Hence find the total cable used for the first

5

displays and the limiting total if the process continues indefinitely. State one limitation of the model.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Segments in stage $n$ : $6\times2^{n-1}$ Cable in stage $n$ : $54\left(\dfrac23\right)^{n-1}$ m First $5$ displays: $\dfrac{422}{3}$ m Limiting total $=162$ m For example, segments eventually become too short to manufacture.	6
Notes
At each stage the number of segments doubles, so stage $n$ has $6\times2^{n-1}$ segments. Each segment then has length $9(1/3)^{n-1}$ metres, giving total stage length $54(2/3)^{n-1}$ metres. The first five displays use $54[1-(2/3)^5]/(1-2/3)=422/3$ metres. Since $\|2/3\|<1$ , the limiting total is $54/(1-2/3)=162$ metres. Indefinite continuation is unrealistic because the segments eventually become impractically short.

(6 marks)

Tier 3 · Hard

The concentration

C_n

milligrams per litre of a pollutant

n

days after monitoring begins is modelled by

C_n=ar^n

, where

a>0

and

0<r<1

. Measurements give

C_2=72

and

C_5=9

. Find

a

and

r

. Find the first integer value of

n

for which the model predicts

C_n<1

. State one limitation of the model.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$a=288$ , $r=\dfrac12$ $n=9$ For example, the model assumes the same proportional decrease continues each day.	6
Notes
The measurements give $ar^2=72$ and $ar^5=9$ . Dividing gives $r^3=1/8$ , so $r=1/2$ , and then $a=72/(1/2)^2=288$ . The condition is $288(1/2)^n<1$ , or $2^n>288$ . Since $2^8=256$ and $2^9=512$ , the first integer value is $n=9$ . The model ignores changes in removal rate, further pollution and any lower physical detection limit.

(6 marks)

Q10

Tier 3 · Hard

10.

A greenhouse model records a daytime temperature

D_n

and the following night-time temperature

N_n

, in degrees Celsius. It uses

D_1=20

N_n=0.6D_n+2

and

D_{n+1}=0.5N_n+9

. Show that

D_{n+1}=0.3D_n+10

. Find an exact formula for

D_n

. Hence find the limiting daytime and night-time temperatures predicted by the model. State one limitation of the model.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$D_n=\dfrac{100}{7}+\dfrac{40}{7}\left(\dfrac{3}{10}\right)^{n-1}$ Limiting daytime temperature $=\left(\dfrac{100}{7}\right)^\circ\mathrm{C}$ Limiting night-time temperature $=\left(\dfrac{74}{7}\right)^\circ\mathrm{C}$ For example, the model assumes the same heating and cooling rules every day.	7
Notes
Substitution gives $D_{n+1}=0.5(0.6D_n+2)+9=0.3D_n+10$ . Its fixed value $L$ satisfies $L=0.3L+10$ , so $L=100/7$ . Therefore $D_{n+1}-100/7=(3/10)(D_n-100/7)$ . Since $D_1-100/7=40/7$ , it follows that $D_n=100/7+(40/7)(3/10)^{n-1}$ . As $\|3/10\|<1$ , $D_n$ tends to $100/7$ . Then $N_n=0.6D_n+2$ tends to $(3/5)(100/7)+2=74/7$ . Real greenhouse temperatures would also depend on weather, ventilation and changing heater performance.

(7 marks)

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