D6 - Sequence and series models | AQA A-Level Mathematics

Translate repeated changes into arithmetic, geometric or recurrence models. Define the index and update order, distinguish an individual amount from a cumulative total, and interpret predictions through their assumptions and physical limits.

Selecting a model and defining its index

An account starts with $£1000$ , earns $1\%$ each month and receives a monthly deposit of $£100$ . Can the same rate and deposit produce two different balances after two months? Yes, if the deposit timing differs. Before calculating, a model must say when each change occurs.

For a simpler comparison, a tank loses either $20$ litres per hour or $4\%$ of its current contents per hour. At an initial $500$ litres, both rules remove $20$ litres in the first step. Their later predictions differ.

You already know

An arithmetic sequence adds a fixed amount. A geometric sequence multiplies by a fixed ratio. Their nth terms must be recalled; their finite sum formulae are supplied in the formula booklet.

Let the index name an observation: “after $n$ complete hours” starts at $n=0$ , while “the payment in month $n$ ” usually starts at $n=1$ . A recurrence combines operations when neither a fixed difference nor a fixed ratio alone describes the update.

Discrete model

A discrete model represents quantities at separate indexed steps. Define the quantity, its units, the starting index and the update rule. A term represents one observation or contribution; a series adds contributions. Adding repeated stock balances usually counts the same stock more than once.

Choose the rule from the stated mechanism. Similar first steps do not make two models interchangeable.

Worked example 1: fixed removal

Question. A tank initially contains $500$ litres. At each hour's end it loses $20$ litres, with no inflow. Model the amount after $n$ complete hours and find the amount after two hours. State the physically admissible integer range before the rule must change.

Plan. Start the index at zero and subtract the same amount per update.

\begin{aligned} V_n&=500-20n && \text{initial amount minus }n\text{ removals}\\ V_2&=500-40=460 && \text{two complete hours}\\ 500-20n&\ge0\ \Longrightarrow\ n\le25 && \text{require non-negative volume} \end{aligned}

Why this step? The starting amount is $V_0$ ; using $n-1$ would delay the first removal by an hour.

Answer. $V_n=500-20n$ litres for $n=0,1,\ldots,25$ , with $460$ litres after two hours.

Check. $V_{25}=0$ ; continuing the unmodified rule gives an impossible negative volume.

Change only the removal mechanism to a fixed percentage. The first removal agrees, but later removals use the new current amount.

Worked example 2: proportional removal

Question. A tank initially contains $500$ litres. At each hour's end it loses $4\%$ of its current contents, with no inflow. Model the amount after $n$ complete hours and find the amount after two hours.

Plan. Use the retained proportion $0.96$ once per completed interval.

\begin{aligned} V_n&=500(0.96)^n && \text{multiply once per hour}\\ V_1&=480 && \text{first removal}\\ V_2&=480(0.96)=460.8 && \text{remove }4\%\text{ of the new amount} \end{aligned}

Why this step? The second loss is $0.04(480)=19.2$ litres, rather than the fixed $20$ litres in example 1.

Answer. $V_n=500(0.96)^n$ litres for integer $n\ge0$ ; $V_2=460.8$ litres.

Check. Both models give $480$ litres after one hour, but only the percentage model keeps multiplying the current amount.

These context-specific formulas are not in the formula booklet; construct them from the stated changes. Hourly production rising from $80$ items by $5$ items each hour similarly gives $u_n=80+5(n-1)$ , with $u_5=100$ . A rise of $5\%$ instead gives $80(1.05)^{n-1}$ . The rule must match what “more” means.

Watch out

Wrong: subtract $4\%$ of the original amount at every step in a current-amount model. Right: use a multiplier of $0.96$ on each new balance. State the index before choosing the exponent.

Use a new fixed-removal situation and check its physical domain.

Arithmetic payments and cumulative totals

A monthly payment is one term. The total paid over several months is a sum of distinct payments. Keep the symbols separate so that a savings target is tested against the total, not against the latest contribution.

For contributions starting at $£50$ and increasing by $£10$ , the recalled nth term is $u_n=50+10(n-1)$ . The supplied arithmetic sum gives $S_n=5n(n+9)$ . These particular expressions are not printed in the formula booklet; derive them from the model.

Worked example 3: reaching a savings target

Question. A club saves $£50$ in month one and $£10$ more each later month. Find the first month by whose end its total reaches at least $£1000$ . Assume no interest, withdrawals or missed payments.

Plan. Model the cumulative sum and check consecutive integer counts around the target.

\begin{aligned} S_n&=\frac n2[100+10(n-1)]=5n(n+9) && \text{sum distinct contributions}\\ S_{10}&=5(10)(19)=950 && \text{test the lower count}\\ S_{11}&=5(11)(20)=1100 && \text{test the next count} \end{aligned}

Why this step? All contributions are positive, so the cumulative sum increases; consecutive boundary checks establish the first successful month.

Answer. Month $11$ .

Check. The eleventh contribution is $£150$ , taking $£950$ to $£1100$ . The target need not be reached exactly.

Change only the savings target to $£2000$ . Equality now occurs at an integer count, so “at least” must include it.

Worked example 4: a target reached exactly

Question. For the same club contributions, find the first month by whose end the total reaches at least $£2000$ .

Plan. Use the same increasing total and bracket the new target.

\begin{aligned} S_{15}&=5(15)(24)=1800 && \text{check the preceding count}\\ S_{16}&=5(16)(25)=2000 && \text{check the target count} \end{aligned}

Why this step? “At least” includes equality, so there is no need to wait until month $17$ .

Answer. Month $16$ .

Check. The sixteenth contribution is $£200$ , exactly the gap from $£1800$ to $£2000$ .

The model assumes the rising contributions remain affordable and are made as stated. Interest would require a different total because older payments would have more time to grow. An arithmetic sum also counts theatre seats: $12$ rows starting at $18$ seats and increasing by $2$ per row have last row $40$ and total $12(18+40)/2=348$ seats.

Watch out

Wrong: compare a cumulative savings target with the single payment $u_n$ . Right: compare a cumulative target with $S_n$ , then check the neighbouring integer counts. A term and a sum measure different things.

Apply the same boundary strategy to a different savings plan.

Percentage changes and cumulative effects

A $20\%$ decrease retains $80\%$ , so its multiplier is $0.8$ . After $n$ complete intervals, a pure proportional model has $V_n=V_0r^n$ . Percentage multipliers and this zero-based expression are not separately supplied in the formula booklet; construct them from the stated change and the recalled geometric nth term.

Equal percentage rises and falls do not undo each other. Starting at $100$ , a $10\%$ rise gives $110$ ; a $10\%$ fall then gives $99$ . The second percentage acts on a different base. The combined multiplier is $1.1(0.9)=0.99$ .

For distinct stage distances, a geometric sum is appropriate. Suppose the first stage covers $10$ metres and each later stage covers $3/4$ of the preceding distance.

Worked example 5: one stage's distance

Question. Under this stage-distance model, find the distance covered in stage four.

Plan. Treat a stage distance as one geometric term, with stage one as the first term.

\begin{aligned} u_n&=10\left(\frac34\right)^{n-1} && \text{define the stage index}\\ u_4&=10\left(\frac34\right)^3=\frac{135}{32} && \text{three reductions} \end{aligned}

Why this step? The question asks for one stage, so use a term rather than the total distance across stages.

Answer. $135/32=4.21875$ metres.

Check. The first four distances are $10,7.5,5.625,4.21875$ metres; each is three quarters of the previous one.

Change only the requested quantity to the total over the first four stages. Those stage distances are distinct contributions, so they can be added.

Worked example 6: the distance across four stages

Question. Find the total distance over the first four stages of the same model.

Plan. Use the supplied finite geometric sum with four terms.

\begin{aligned} S_4&=\frac{10[1-(3/4)^4]}{1-3/4} && \text{sum four distinct distances}\\ &=\frac{875}{32}=27.34375 && \text{evaluate exactly} \end{aligned}

Why this step? Summing the four distances answers “in total”; reporting $u_4$ would give only the final stage's distance.

Answer. $27.34375$ metres.

Check. $10+7.5+5.625+4.21875=27.34375$ .

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Pure Mathematics

Mechanics

Statistics

D6 - Sequence and series models

Selecting a model and defining its index

Arithmetic payments and cumulative totals

Percentage changes and cumulative effects

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