Sequences and series in modellingEdexcel A-Level Maths: Revision notes

Read the situation and decide how the quantity changes from one period to the next.

• Increases by the same amount each period: arithmetic, $u_n=a+(n-1)d$.
• Increases or decreases by the same percentage each period: geometric, $u_n=ar^{n-1}$. A rise of $p\%$ gives $r=1+\frac{p}{100}$; a fall of $p\%$ gives $r=1-\frac{p}{100}$.
• Defined by a rule from the previous value: a recurrence relation, $u_{n+1}=f(u_n)$.

Write down $a$ and $d$ or $r$ first, and say which model you are using.

Aisha saves £50 in month 1 and £8 more each month: $a=50$, $d=8$. The 12th payment is $50+11\times8=£138$ and the total over 12 months is $S_{12}=\frac{12}{2}[2(50)+11(8)]=£1128$. To find when a payment first exceeds £300, solve $50+8(n-1)>300$ to get $n=33$.

Always check the answer in context: months are whole numbers, so round an inequality solution up to the next whole month.

A tree grows $0.50$ m in year 1 and $85\%$ of the previous year's growth thereafter: growths $0.5,0.425,0.361,\ldots$ form a geometric sequence with $r=0.85$. Total growth over $n$ years is $S_n=\frac{0.5(1-0.85^n)}{0.15}$. Over 5 years that is $1.854$ m, so the height is $1.2+1.854=3.05$ m.

Because $|r|<1$, the growth sum converges: total growth never exceeds $\frac{0.5}{0.15}=3.33$ m, so the model predicts a maximum height of $1.2+3.33=4.53$ m. This is a limit the tree approaches but never reaches.

With 3% interest added at the end of each year, each £1 becomes £1.03. If Ben pays £1500 at the start of each year, the payment made at the start of year 1 earns interest for $n$ years by the end of year $n$, the next for $n-1$ years, and so on. After 3 years:

$1500(1.03^3+1.03^2+1.03)=£4775.44.$

This is a geometric series with first term $1500\times1.03=1545$, ratio $1.03$, so after $n$ years $S_n=\frac{1545(1.03^n-1)}{0.03}$; for $n=10$ that is £17 711.69.

Arithmetic growth is linear and geometric growth is exponential, so a geometric plan can start behind and later overtake. Plan A: £100 then £12 more each month. Plan B: £100 then 8% more each month. Plan B's payment first exceeds plan A's in month 12 ($233.16$ against $232$). Over 24 months plan B's total is $\frac{100(1.08^{24}-1)}{0.08}=£6676$ against plan A's £5712.

For a payment target, solve $a+(n-1)d>T$ directly. For a geometric target use logs: $ar^{n-1}>T$ gives $n-1>\frac{\ln(T/a)}{\ln r}$ for $r>1$. When a formula cannot be solved neatly (comparing two different types of growth), test successive values of $n$ in a table.

A model can be defined by a rule such as $b_{n+1}=1.01b_n-200$, a loan balance with 1% interest added monthly and £200 repaid. With $b_1=5000$: $b_2=4850$, $b_3=4698.50$. Calculate terms in order and describe the pattern.

Models are simplifications. A geometric model of growth cannot continue for ever in reality (resources run out), and a constant interest rate or fixed saving may not hold. A good comment names the assumption that fails, e.g. 'interest rates may change over ten years'.