1.04k - Modelling growth, decay and finance with sequences | OCR A-Level Mathematics A

Choose a discrete model for equal changes or percentage changes, and use it for interest, savings and repayments. You will interpret integer-time thresholds, payment timing and the limits of a financial prediction.

Model equal changes or equal multipliers

Can you make regular loan repayments forever without reducing the debt at all? A $1000$ -pound balance charged $2\%$ per month gains $20$ pounds in interest immediately. Paying exactly $20$ pounds returns it to $1000$ . We will find what a larger payment changes, but first separate addition from multiplication.

You already know

An AP adds a fixed amount; a GP multiplies by a fixed factor. An increase of $p\%$ multiplies by $1+p/100$ , while a decrease multiplies by $1-p/100$ .

A $10\%$ rise followed by a $10\%$ fall does not cancel: $1.1(0.9)=0.99$ , leaving a $1\%$ loss. Percentages act on the current quantity, not always the starting quantity.

Discrete change models

If $X_0$ is the initial value and $n$ counts completed intervals,

X_n=X_0+nd\quad\text{or}\quad X_n=X_0q^n

The first adds constant $d$ ; the second uses constant multiplier $q$ . Here $n$ is a non-negative integer.

Not in the formula sheet: learn it.

The exponent is $n$ , not $n-1$ , because the initial observation is indexed zero.

Worked example 1: equal additions

Question. A stock starts at $200$ units and increases by $30$ units each week. Find its value after three weeks and state an assumption.

Plan. Count completed weeks from zero and add the same amount each time.

\begin{aligned} X_n&=200+30n && \text{constant weekly change} \\ X_3&=200+90=290 && \text{three intervals} \end{aligned}

Why start at zero? Before any week passes, the stock is already $200$ .

Answer. $290$ units, assuming each week's net addition stays at $30$ units.

Check. The successive values are $200,230,260,290$ .

Change only the rule for the weekly change: it is now $15\%$ of the current stock, equal to $30$ units only in the first week.

Worked example 2: equal multipliers

Question. The same initial stock grows by $15\%$ per week. Find its value after three weeks and state an assumption.

Plan. Apply the weekly retention-and-growth factor three times.

\begin{aligned} q&=1+0.15=1.15 && \text{percentage multiplier} \\ X_n&=200(1.15)^n && \text{geometric model} \\ X_3&=304.175 && \text{evaluate unrounded} \end{aligned}

Why is the answer above $290$ ? After week one, the same percentage acts on a larger stock.

Answer. The model predicts about $304$ units, assuming a constant weekly percentage with no other gains or losses.

Check. The first values are $200,230,264.5,304.175$ .

Data $100,120,140$ suggest constant differences, while $100,120,144$ suggest constant ratios. A short fit does not establish that a rule will persist. Compare later observations and restrict the model to a sensible time range.

Watch out

Wrong: losing $12\%$ means multiplying by $0.12$ . Right: $88\%$ remains, so multiply by $0.88$ . State the interval and the meaning of the index.

Try a percentage loss, changing the multiplier's direction.

Interest changes the principal or the whole balance

Simple interest is calculated on the initial principal. Compound interest is calculated on the current balance, so earlier interest starts earning interest too.

Principal and interest rate

The principal $P$ is the amount initially deposited or borrowed. A rate $i$ per interval is written as a decimal: $4\%$ means $i=0.04$ . The interval must match the interest-crediting rule.

No deposits or withdrawals are included in the next two formulas.

Simple and compound balances

For $n$ completed intervals at fixed rate $i$ ,

B_n=P(1+ni)\quad\text{(simple)},\qquad B_n=P(1+i)^n\quad\text{(compound)}

These assume the stated interest rule, no fees or tax and no other transactions.

Not in the formula sheet: learn it.

The first formula is arithmetic; the second is geometric.

Worked example 3: interest on the principal

Question. Find the balance after three years for $5000$ pounds at $4\%$ annual simple interest.

Plan. Find the fixed annual interest and add three copies.

\begin{aligned} Pi&=5000(0.04)=200 && \text{annual interest} \\ B_3&=5000+3(200)=5600 && \text{three equal additions} \end{aligned}

Why is the annual interest fixed? Simple interest uses the initial $5000$ each year.

Answer. $5600$ pounds.

Check. Total interest is $600$ pounds, or $12\%$ of the principal.

Change only the interest rule from simple to annually compounded.

Worked example 4: interest on earlier interest

Question. Find the corresponding compound balance after three years and compare it with simple interest.

Plan. Multiply the balance by $1.04$ three times, then subtract the simple balance.

\begin{aligned} B_3&=5000(1.04)^3=5624.32 && \text{compound balance} \\ 5624.32-5600&=24.32 && \text{extra interest} \end{aligned}

Why is there an extra amount? Interest credited in years one and two itself earns interest later.

Answer. $5624.32$ pounds, $24.32$ pounds more than simple interest.

Check. The year-end balances are $5200$ , $5408$ and $5624.32$ pounds.

A monthly rate of $0.5\%$ gives annual growth factor $1.005^{12}\approx1.06168$ , not exactly $1.06$ . Conversely, an effective annual rate cannot simply be divided by twelve to obtain an equivalent effective monthly rate. Use the convention supplied, keep guard digits and round a final money amount to pennies when requested.

Watch out

Wrong: substitute a monthly rate with a number of years. Right: match the rate and interval, so three years at a stated monthly rate means $36$ intervals.

Change the rate and compare the two rules again.

Logarithms locate an integer-time threshold

When the unknown is an exponent, logarithms expose it. You need only a small bridge here; fuller logarithm methods come later.

For positive $z$ , $\log z$ means the power of $10$ giving $z$ : $10^{\log z}=z$ . The function is increasing, so taking logarithms preserves inequalities between positive quantities.

Logarithms of positive powers

For $q>0$ ,

\log(q^n)=n\log q

To see why, write $q=10^{\log q}$ , so $q^n=10^{n\log q}$ . Natural logarithms give the same threshold quotient when used consistently.

Not in the formula sheet: learn it.

If $0<q<1$ , then $\log q<0$ , and division by it reverses an inequality. For instance, $200(0.8)^n<50$ gives

n\log0.8<\log0.25\quad\Longrightarrow\quad n>\frac{\log0.25}{\log0.8}\approx6.21257

The first allowed integer is $7$ : the values at $6$ and $7$ are $52.4288$ and $41.94304$ . This is a ceiling decision, not ordinary rounding.

Worked example 5: first crossing in growth

Question. Determine the first whole year for which $1000(1.06)^n>1500$ .

Plan. Divide out the initial balance, take logarithms, then check neighbouring integers.

\begin{aligned} 1.06^n&>1.5 && \text{divide by }1000 \\ n\log1.06&>\log1.5 && \text{positive arguments} \\ n&>6.95851\ldots && \text{divide by positive }\log1.06 \end{aligned}

Why choose $7$ ? It is the smallest integer strictly above the calculated boundary.

Answer. The first whole year is year $7$ .

Check. Year six gives $1418.52$ pounds; year seven gives $1503.63$ pounds, to pennies.

Change only which side of the same boundary is wanted: ask for the last year below the target.

Worked example 6: last year before crossing

Question. Determine the last non-negative whole year for which $1000(1.06)^n<1500$ .

Plan. Solve the reversed inequality and choose the largest permitted integer.

\begin{aligned} 1.06^n&<1.5 && \text{normalise} \\ n&<\frac{\log1.5}{\log1.06}=6.95851\ldots && \text{positive log divisor} \end{aligned}

Why not select $7$ again? The question now asks for a value before the crossing.

Answer. Year $6$ is the last whole year below $1500$ pounds.

Check. The same two model values straddle the boundary, and the balance increases each year.

For a strict boundary exactly at $n=6$ , $n>6$ starts at $7$ , whereas $n\ge6$ includes $6$ . Logarithmic comparisons, such as $\log(X_n/200)<\log0.25$ , are equivalent to comparing the positive arguments. Zero or negative arguments are invalid; $q=1$ gives a constant model and must be checked directly because $\log1=0$ .

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1.04k - Modelling growth, decay and finance with sequences

Model equal changes or equal multipliers

Interest changes the principal or the whole balance

Logarithms locate an integer-time threshold

Continue this lesson.

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