Modelling with sequences and seriesAQA A-Level Maths: Revision notes

To model with a sequence, decide how each term relates to the previous one.

• A constant amount added each time (a fixed rise in profit, equal increases in payments) is arithmetic: $u_n=a+(n-1)d$.
• A constant percentage or multiplier (interest, a rebound ratio, depreciation) is geometric: $u_n=ar^{n-1}$, with $r=1+\frac{p}{100}$ for a $p\%$ rise or $r=1-\frac{p}{100}$ for a fall. Define what $u_1$ and $n$ stand for in context, for example 'profit in year $n$'.

A profit of £$40\,000$ rising by £$5000$ a year is arithmetic with $a=40\,000$, $d=5000$.

• Profit in year $10$: $40\,000+9\times5000=85\,000$.
• Total over $10$ years: $\frac{10}{2}(80\,000+45\,000)=625\,000$.
• First year over £$200\,000$: $40\,000+5000(n-1)>200\,000$ gives $n>33$, so year $34$ (year $33$ is exactly £$200\,000$).

A ball dropped from $2$ m that rebounds to $70\%$ of its previous height rises to $2(0.7)^n$ after $n$ bounces: $1.4,0.98,0.686,\dots$ Total distance until rest: the initial drop plus every rise and every fall, so $2+2\times\frac{1.4}{1-0.7}=\frac{34}{3}\approx11.3$ m. The $2\times$ is because each rise is matched by a fall. Compound interest: £$1000$ invested at the start of each year at $4\%$. The deposit made at the start of year $k$ is worth $1000(1.04)^{n-k+1}$ at the end of year $n$, so the total is $1000(1.04)+1000(1.04)^2+\dots+1000(1.04)^n$, a geometric series with $a=1040$, $r=1.04$: $S_n=\frac{1040(1.04^n-1)}{0.04}$. After $5$ years this is £$5633$.

To find when a total passes a target, write $S_n$ as a function of $n$, form an inequality, and solve.

• Arithmetic: a quadratic in $n$. A loan of £$11\,800$ with payments $400,420,\dots$ gives $10n^2+390n=11\,800$, so $n^2+39n-1180=0$, $(n-20)(n+59)=0$, $n=20$.
• Geometric: take logarithms. $26\,000(1.04^n-1)>20\,000$ gives $1.04^n>1.769$ and $n>14.5$, so $n=15$. Reject impossible roots (negative $n$), and give a whole number. For the loan, the final payment is $400+19\times20=780$, the last $5$ payments sum to $S_{20}-S_{15}=3700$, and the mean payment is $\frac{11\,800}{20}=590$.

Always put your answer back in context: units (£, metres, months), rounding (whole years or months) and what the number means. Comment on assumptions when asked. An arithmetic profit model rises forever, so it cannot hold in the long run. A bouncing ball model assumes the same ratio at every bounce. A savings model assumes a constant rate of interest and deposits made on time. A geometric model with $|r|<1$ has a finite limit (the sum to infinity), which can represent a total distance or total quantity.