Sequences and series in mathematical modelling

A sequence model describes a quantity measured at regular stages. A series model adds those stage values to find a cumulative total.

The central question is not initially which formula to use. It is what changes from one stage to the next:

Wording or structure	Mathematical change	Likely model
increases by £250 each year	add $250$	arithmetic sequence
decreases by $18$ units each cycle	subtract $18$	arithmetic sequence
increases by $6\%$ each year	multiply by $1.06$	geometric sequence
retains $82\%$ of its value each year	multiply by $0.82$	geometric sequence
total produced over several weeks	add weekly amounts	series

Real data are rarely exactly arithmetic or geometric. A model is a deliberate simplification whose predictions are useful only while its assumptions remain reasonable.

Prerequisites

You should be able to:

• find terms and sums of arithmetic sequences and series;
• find finite and infinite sums of geometric sequences and series;
• convert a percentage change into a multiplier;
• solve exponential equations using logarithms.

A reliable modelling process

For each problem:

1. Define what $u_n$ represents, including its unit and time.
2. Decide where $n=1$ occurs.
3. Translate the repeated change into $d$ or $r$.
4. Decide whether the question asks for one term $u_n$ or a sum $S_n$.
5. Calculate, keeping full precision until the end.
6. Interpret the result and check its domain, units and realism.

The second step prevents many errors. If £800 is deposited at the start of year 1, it has earned interest by the end of that year. If it is deposited at the end of year 1, it has not.

Constant amount is not constant percentage

An increase of $5\%$ is not the same as adding the same number each time. Five per cent is calculated from the current value, so the added amount changes. It produces multiplication by $1.05$ and therefore a geometric model.

Constant additive change

If a quantity starts at $a$ and changes by the fixed amount $d$ per stage, then

$u_n=a+(n-1)d.$

If each stage’s amount contributes to a total, then

$S_n=\frac n2\bigl(2a+(n-1)d\bigr).$

Worked example 1: one value and a cumulative total

A factory makes $1250$ components in week 1. Its weekly output rises by $80$ components each week.

Find the output in week 12 and the total output during the first 12 weeks.

The weekly outputs form an arithmetic sequence with

$a=1250,\qquad d=80.$

The week 12 output is one term:

$$u_{12}=1250+11(80)=2130.$$

The total during the 12 weeks is a sum:

$$\begin{aligned} S_{12} &=\frac{12}{2}\bigl(2(1250)+11(80)\bigr)\\ &=6(3380)\\ &=20280. \end{aligned}$$

Therefore the factory makes $2130$ components in week 12 and $20280$ components altogether. The first answer has units of components per week; the second has units of components.

Worked example 2: interpreting an arithmetic model

A machine contains $640$ ml of lubricant immediately after servicing. A model assumes that it loses $24$ ml per day. Let $u_n$ be the amount immediately after $n-1$ complete days.

Then

$u_n=640-24(n-1).$

Find the first value of $n$ for which the model predicts less than $100$ ml.

Solve

$640-24(n-1)<100.$

Thus

$$-24(n-1)<-540,$$

and reversing the inequality when dividing by $-24$ gives

$n-1>22.5,\qquad n>23.5.$

The first integer value is $n=24$. This represents the amount after $23$ complete days:

$u_{24}=640-24(23)=88.$

The index is not the elapsed time. By definition, $u_1$ is at time $0$, so $u_{24}$ is at time $23$ days.

The model cannot sensibly continue after the lubricant reaches zero. It also assumes a constant loss regardless of temperature, machine use and the remaining volume.

Self-check 1

A hall has $28$ seats in its first row and each later row has $4$ more seats than the previous row. There are $18$ rows.

1. How many seats are in row 18?
2. How many seats are there altogether?
3. State one assumption of the model.

Answers

Here $a=28$, $d=4$ and $n=18$.

$u_{18}=28+17(4)=96.$

$S_{18}=\frac{18}{2}(28+96)=1116.$

A suitable assumption is that every row has exactly four more usable seats than the row before. Real obstructions or aisle spaces could make this false.

Constant proportional change

If a quantity begins at $a$ and is multiplied by the fixed factor $r$ at every stage, then

$u_n=ar^{n-1}.$

A percentage increase of $p\%$ gives

$r=1+\frac p{100},$

whereas a percentage decrease of $p\%$ gives

$r=1-\frac p{100}.$

Worked example 3: depreciation and time indexing

A new machine is worth £48 000. Its value is modelled as decreasing by $14\%$ at the end of each year. Find its modelled value after 5 years.

It retains

$100\%-14\%=86\%$

of its value each year, so $r=0.86$.

If $V_n$ denotes the value after $n$ years, then

$V_n=48000(0.86)^n.$

Therefore

$V_5=48000(0.86)^5\approx£22580.50.$

Why is the exponent $5$, rather than $4$? The initial £48 000 is the value at time $0$. Five annual reductions have occurred after 5 years.

If instead the first term were defined as $u_1=48000$, then $u_n=48000(0.86)^{n-1}$ and the value after 5 years would be $u_6$. Both descriptions are correct when used consistently.

Worked example 4: find when a threshold is crossed

A colony initially contains $350$ bacteria and grows by $18\%$ every hour. According to the model, after how many complete hours will the colony first exceed $2000$ bacteria?

After $t$ hours,

$N_t=350(1.18)^t.$

We need

$350(1.18)^t>2000.$

Hence

$1.18^t>\frac{40}{7}.$

Taking logarithms gives

$$t>\frac{\log(40/7)}{\log(1.18)}\approx10.53.$$

The least whole number of complete hours is $11$. Check the boundary:

$N_{10}\approx1832<2000,\qquad N_{11}\approx2162>2000.$

Do not simply round $10.53$ to the nearest integer. The word first requires the smallest integer satisfying the inequality.

The model assumes an unchanged percentage growth rate and no limiting effects such as shortage of nutrients or space. Long term use would be unrealistic.

When the cumulative total is geometric

Suppose the amount during stage $n$ is

$u_n=ar^{n-1}.$

The total during the first $n$ stages is

$S_n=\frac{a(1-r^n)}{1-r},\qquad r\ne1.$

The term $u_n$ answers questions such as “how much in year $n$?” The sum $S_n$ answers questions such as “how much over the first $n$ years?”

Worked example 5: total production under percentage growth

A solar array generates $9200$ kWh in its first year. Its annual output is modelled as decreasing by $2.5\%$ each year.

Find its output in year 8 and its total output during its first 8 years.

The multiplier is

$r=1-0.025=0.975.$

The year 8 output is

$u_8=9200(0.975)^7\approx7705.84\text{ kWh}.$

The total is

$$\begin{aligned} S_8 &=\frac{9200(1-0.975^8)}{1-0.975}\\ &\approx67472.14\text{ kWh}. \end{aligned}$$

The exponent in $u_8$ is $7$, but the finite sum contains $r^8$. These are different formulae answering different questions.

Worked example 6: regular payments with interest

At the start of each year, Priya deposits £600 into an account. Interest of $4\%$ is added at the end of each year. Find the value of the first five deposits immediately after interest is added at the end of year 5.

Draw the growth history of each deposit:

Deposit made	Number of interest additions by end of year 5	Value then
start of year 1	$5$	$600(1.04)^5$
start of year 2	$4$	$600(1.04)^4$
start of year 3	$3$	$600(1.04)^3$
start of year 4	$2$	$600(1.04)^2$
start of year 5	$1$	$600(1.04)$

The account value is

$600(1.04)+600(1.04)^2+\cdots+600(1.04)^5.$

This is a five term geometric series with first term $600(1.04)$ and ratio $1.04$:

$$\begin{aligned} A &=\frac{600(1.04)(1.04^5-1)}{1.04-1}\\ &\approx£3379.79. \end{aligned}$$

If deposits were made at the end of each year, the final deposit would earn no interest before the valuation time. The powers would then run from $0$ to $4$. A timeline is safer than memorising a special formula.

Self-check 2

A runner covers $3$ km in week 1 and increases the weekly distance by $7\%$ each week.

1. Find the distance in week 10.
2. Find the total distance during the first 10 weeks.
3. Find the first week in which the weekly distance exceeds $6$ km.

Answers

Here $a=3$ and $r=1.07$.

$u_{10}=3(1.07)^9\approx5.52\text{ km}.$

$S_{10}=\frac{3(1.07^{10}-1)}{0.07}\approx41.45\text{ km}.$

For the threshold,

$3(1.07)^{n-1}>6,$

so

$n-1>\frac{\log2}{\log1.07}\approx10.24.$

Thus $n>11.24$, and the first possible integer is $n=12$. Indeed, $u_{11}\approx5.90$ and $u_{12}\approx6.31$.

Infinite series as limiting models

An infinite geometric model has a finite total only if

$|r|<1.$

Then

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

The answer is the limit approached by partial sums. It does not mean infinitely many physical events have finished.

Worked example 7: total distance travelled by a bouncing ball

A ball is dropped from a height of $3$ m. After every impact, it rebounds to $70\%$ of the previous height. Find the total vertical distance predicted by the model.

The initial drop contributes $3$ m once. Every rebound height is travelled twice, once upwards and once downwards. The rebound heights are

$3(0.7),\quad 3(0.7)^2,\quad 3(0.7)^3,\ldots$

Therefore

$$\begin{aligned} D &=3+2\left(3(0.7)+3(0.7)^2+\cdots\right)\\ &=3+2\left(\frac{2.1}{1-0.7}\right)\\ &=17\text{ m}. \end{aligned}$$

A common incorrect answer is $10$ m, obtained from $3/(1-0.7)$. That counts one direction for each height and misses the separate upward and downward journeys.

The mathematical model permits infinitely many bounces. A real ball eventually deforms, loses energy in a non-constant way and becomes effectively stationary. The limiting distance can still be a useful approximation.

Worked example 8: determine a parameter from a limiting total

A treatment delivers $40$ mg initially. Each later dose is $r$ times the preceding dose, where $0<r<1$. The total amount delivered over all doses is intended to be $160$ mg. Find $r$.

Using the sum to infinity,

$160=\frac{40}{1-r}.$

Therefore

$1-r=\frac14,$

so

$\boxed{r=\frac34}.$

The given restriction confirms convergence and excludes ratios that would make later doses negative.

Choosing and criticising a model

A strong modelling answer connects algebra to the context.

Additive or multiplicative?

Suppose a town has population $P$.

• “Gains 800 residents each year” gives $P_{n+1}=P_n+800$, an arithmetic model.
• “Grows by $4\%$ each year” gives $P_{n+1}=1.04P_n$, a geometric model.

The first adds the same number. The second adds a number proportional to the current population.

Discrete or continuous?

A sequence changes at separate stages such as months or payments. It is most natural when values are measured or updated periodically. A continuous process may be better represented by an exponential growth or decay model.

What makes a model questionable?

Check whether:

• the change really remains constant;
• external conditions remain stable;
• the quantity has a natural upper or lower bound;
• fractional values make sense in context;
• the model is being extrapolated far beyond the observed interval;
• rounding at each stage changes later results.

For example, constant percentage population growth cannot continue indefinitely because resources are finite. Constant depreciation can eventually predict a value below a realistic scrap value. Constant additive decay can predict a negative physical quantity.

Calculator values and real stage values

Unless the context explicitly says that values are rounded at each stage, use unrounded calculator values throughout and round only the final answer. Repeated premature rounding changes the model.

Mixed exam-style check

A woodland contains $12000$ tonnes of usable timber at the start of year 1. During year 1, $900$ tonnes are harvested. The amount harvested in each later year is $96\%$ of the amount harvested in the preceding year.

1. Find the amount harvested in year 8.
2. Find the total harvested during the first 8 years.
3. Assuming the model continues indefinitely, find the total amount harvested.
4. Explain why the model does not predict that all $12000$ tonnes are harvested.

Answer

The annual harvests form a geometric sequence with $a=900$ and $r=0.96$.

$u_8=900(0.96)^7\approx676.30\text{ tonnes}.$

$$S_8=\frac{900(1-0.96^8)}{1-0.96}\approx6268.73\text{ tonnes}.$$

Since $|0.96|<1$,

$S_\infty=\frac{900}{1-0.96}=22500\text{ tonnes}.$

This exceeds the initial $12000$ tonnes, so the numerical result exposes a missing feature rather than proving that the model is sensible. The model may implicitly require regrowth, but no regrowth rate is specified. Without regrowth, it must stop when the available timber is exhausted. A mathematically convergent series can still be physically invalid.

Common misconceptions

• Confusing term and total: use $u_n$ for one stage and $S_n$ for the accumulated amount.
• Starting at the wrong time: state whether the initial value is at time $0$ or during stage $1$.
• Using $1-p$ with a percentage: for a decrease of $p\%$, use $1-p/100$, not $1-p$.
• Rounding a threshold: solve the inequality and test neighbouring integers. Do not automatically round to the nearest integer.
• Forgetting repeated journeys: in rebound problems, most heights are travelled both upwards and downwards.
• Using a sum to infinity without convergence: check $|r|<1$ before using $a/(1-r)$.
• Trusting an impossible prediction: always compare the output with physical bounds and the assumptions of the context.

What to learn next

Use sigma notation to represent more complicated totals compactly. Study exponential growth and decay for continuous proportional change, or sequences and recurrence relations for models in which each new value depends on previous values in a more general way.