1.4 Financial applications of geometric sequences (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Compound interest as a geometric sequence

With compound interest, interest is added to the balance and then earns interest itself. Each period the balance is multiplied by the same factor, so the balances form a geometric sequence. The formula booklet gives $FV = PV\times\left(1+\frac{r}{100k}\right)^{kn}$ where $PV$ is the present value, $FV$ the future value, $r$ the nominal annual rate (%), $n$ the number of years and $k$ the number of compounding periods per year.

5000 USD at 6% compounded annually for 3 years: $5000\times1.06^{3} = 5955.08$ USD.

Key termscompound interestpresent valuefuture value

Common mistake

Using simple interest ( $5000 + 3\times300$ ) when the question says compound. Compound interest multiplies; simple interest adds.

Section 2

Compounding yearly, half-yearly, quarterly or monthly

The value of $k$ depends on how often interest is compounded:

yearly $k = 1$ , half-yearly $k = 2$ , quarterly $k = 4$ , monthly $k = 12$ .

Divide the annual rate by $k$ and multiply the number of years by $k$ . For 6% compounded monthly for 3 years: $5000\left(1+\frac{0.06}{12}\right)^{36} = 5983.40$ USD.

More frequent compounding gives a slightly larger final value for the same nominal rate, because interest starts earning interest sooner. That is why 5.1% compounded monthly can beat 5.2% compounded annually.

Key termsnominal ratecompounding period

Common mistake

Raising to the power $n$ instead of $kn$ : monthly for 3 years needs power 36, not 3.

Exam tip

On the GDC's finance solver (TVM), set $P/Y$ and $C/Y$ to $k$ and enter $PV$ as a negative number (money paid out).

Section 3

Annual depreciation

Depreciation is a loss of value. If an item loses $r\%$ of its value each year, multiply by $1 - \frac{r}{100}$ each year: $\text{value after } n \text{ years} = \text{initial value}\times\left(1-\frac{r}{100}\right)^{n}.$ A 32 000 USD car depreciating 15% a year is worth $32\,000(0.85)^{5} \approx 14\,199$ USD after 5 years.

The value never reaches zero, but it falls quickly at first because each year's loss is 15% of a larger amount.

Key termsdepreciation

Common mistake

Multiplying by 0.15 instead of 0.85 — 0.15 is what is lost, 0.85 is what remains.

Section 4

Inflation and real value

Inflation means prices rise, so money buys less. The real value of an investment measures its purchasing power in today's money.

The usual IB approach is to use a real interest rate $\text{real rate} \approx \text{interest rate} - \text{inflation rate}$ and then compound as normal. For 4.5% interest and 2.8% inflation over 6 years: real value $= 20\,000(1.017)^{6} \approx 22\,129$ EUR, even though the account shows 26 045.20 EUR.

Read the question: if it tells you a different method (for example dividing the future value by $1.028^{6}$ ), use that method.

Key termsinflationreal valuereal interest rate

Exam tip

Always comment in context: a large nominal gain may be a small real gain when inflation is high.

Section 5

Using technology: the TVM solver

Examination questions may require technology, including the GDC's financial package (TVM solver). Enter:

$N$ = total number of compounding periods ( $n\times k$ ), $I\%$ = nominal annual rate,
$PV$ negative (money you pay in), $PMT = 0$ , $FV$ = value received,
$P/Y$ and $C/Y$ = number of compounding periods per year.

Solve for any unknown: the time to double an investment, or the rate needed to reach a target. You will not be asked to derive the compound interest formula.

Key termsTVM solver

Common mistake

Forgetting the sign convention: if PV and FV are both entered as positive the solver returns an error or nonsense.

Must know

$FV = PV\left(1+\frac{r}{100k}\right)^{kn}$ with $k = 1, 2, 4, 12$ for yearly, half-yearly, quarterly, monthly.
Depreciation by $r\%$ a year: multiply by $1 - \frac{r}{100}$ each year.
Real value: use real rate ≈ interest − inflation unless told otherwise.
Time to reach a target: solve with the GDC, then round up to a whole number of periods.
Give money to 2 d.p. or as the question specifies.

That's the notes covered.

Carry on to the next subtopic.