Financial-Applications AI HL P1

Interest Models and Arithmetic Motion

IB Mathematics HL • Financial Mathematics • Simple and Compound Interest • Arithmetic Sequences

⭐ Key Concepts

Simple interest grows linearly with time:

$\mathrm{FV}=\mathrm{PV}(1+rt)$

Compound interest grows exponentially:

$\mathrm{FV}=\mathrm{PV}(1+r)^t$

If interest is compounded $k$ times per year, then

$\mathrm{FV}=\mathrm{PV}\left(1+\frac{r}{k}\right)^{kn}$

where $r$ is written as a decimal.

An arithmetic sequence has constant difference $d$ .
The general term is

$u_n=u_1+(n-1)d$

The sum of the first $n$ terms is

$S_n=\frac{n}{2}\left(2u_1+(n-1)d\right)$

📘 IB Command Terms

Determine – obtain an answer using correct working.
Find – calculate the required value.
Write down – give the required value or expression directly.
Show that – give a clear mathematical argument leading to the stated result.
Hence – use a previous result to obtain the next answer.

🧮 GDC Calculator Tips

Use the power key carefully for compound interest expressions such as $(1.04)^{10}$ .
For financial questions, do not round too early. Keep full calculator values until the final step.
When comparing simple and compound interest, graphing both expressions can help identify the intersection point.
For arithmetic motion questions, store values of $u_n$ and $S_n$ separately to avoid mixing term and sum formulas.

📌 Example 1 — Simple vs Compound Interest

Christian invests $5000 at simple interest of $5\%$ per annum, and Timothy invests $5000 at $4\%$ per annum compounded annually.

(a) Determine how much each will have after $10$ years.

(b) Determine after how many years they will have the same amount.

Solution

(a) Amount after 10 years

Christian uses simple interest:

$\mathrm{FV}=5000(1+0.05\cdot10)$

$=5000(1.5)=7500$

So Christian has

$\boxed{\$7500}$

Timothy uses compound interest:

$\mathrm{FV}=5000(1.04)^{10}$

$\approx 5000(1.48024)=7401.20$

So Timothy has

$\boxed{\$7401.20}$

📊 Marks: M1 A1 M1 A1

(b) Same amount after $t$ years

Set the amounts equal:

$5000(1+0.05t)=5000(1.04)^t$

$1+0.05t=(1.04)^t$

This equation must be solved numerically. Using trial, graphing, or a GDC intersection method gives

$t\approx 11.9$

So they have the same amount after approximately

$\boxed{11.9\text{ years}}$

At $t=12$ :

$1+0.05(12)=1.6$

$(1.04)^{12}\approx 1.601$

This confirms the intersection is very close to $12$ years.

📊 Marks: M1 M1 A1

Quick Check

$6000 is invested at $6\%$ simple interest and another $6000 at $4.5\%$ compounded annually.

(i) Find the amounts after $8$ years.

(ii) Determine when they are equal.

Solution

Simple interest:

$6000(1+0.06\cdot8)=6000(1.48)=8880$

Compound interest:

$6000(1.045)^8\approx 6000(1.411)=8466$

So after $8$ years the amounts are

$\boxed{\$8880}\quad\text{and}\quad\boxed{\$8466}$

For equality:

$1+0.06t=(1.045)^t$

Numerically,

$t\approx 10.8\text{ years}$

📌 Example 2 — Compounding Frequency

Linda invests €2400 at a 3.5\% nominal annual rate. Find the value after $25$ years if interest is compounded:

(a) half-yearly (b) quarterly (c) monthly

Solution

Use

$\mathrm{FV}=\mathrm{PV}\left(1+\frac{r}{k}\right)^{kn}$

with

$\mathrm{PV}=2400,\qquad r=0.035,\qquad n=25$

(a) Half-yearly compounding

$k=2$

$\mathrm{FV}=2400\left(1+\frac{0.035}{2}\right)^{50}$

$=2400(1.0175)^{50}\approx 5714$

$\boxed{€5714}$

📊 Marks: M1 A1

(b) Quarterly compounding

$k=4$

$\mathrm{FV}=2400\left(1+\frac{0.035}{4}\right)^{100}$

$=2400(1.00875)^{100}\approx 5734$

$\boxed{€5734}$

📊 Marks: M1 A1

(c) Monthly compounding

$k=12$

$\mathrm{FV}=2400\left(1+\frac{0.035}{12}\right)^{300}$

$\approx 2400(1.002916\overline{6})^{300}\approx 5750$

$\boxed{€5750}$

Monthly compounding gives the largest final value, as expected.

📊 Marks: M1 A1

Quick Check

€5000 is invested at $4.2\%$ nominal annual interest for $18$ years.

Find the future value if interest is compounded

(i) annually (ii) quarterly (iii) monthly

Solution

(i) Annual compounding

$5000(1.042)^{18}\approx 10510$

$\boxed{€10510}$

(ii) Quarterly compounding

$5000\left(1+\frac{0.042}{4}\right)^{72}\approx 10661$

$\boxed{€10661}$

(iii) Monthly compounding

$5000\left(1+\frac{0.042}{12}\right)^{216}\approx 10717$

$\boxed{€10717}$

📌 Example 3 — Pebble Falling: Arithmetic Sequence of Distances

Emily drops a pebble from a cliff. It falls $5$ m in the first second, $15$ m in the next, $25$ m in the third, and continues similarly.

The per-second distances form an arithmetic sequence.

(a) Write the common difference $d$ .

(b) Find the distance during the $4$ th second.

(c) Find the distance during the $15$ th second.

(d) Find the total distance in the first $15$ seconds, giving your answer in kilometres.

Solution

Given

$u_1=5,\qquad u_2=15$

So the common difference is

$d=15-5=10$

(a)

$\boxed{d=10\text{ m}}$

📊 Marks: M1 A1

(b) Distance during the 4th second

$u_4=u_1+(4-1)d$

$u_4=5+3(10)=35$

$\boxed{35\text{ m}}$

📊 Marks: M1 A1

(c) Distance during the 15th second

$u_{15}=u_1+(15-1)d$

$u_{15}=5+14(10)=145$

$\boxed{145\text{ m}}$

📊 Marks: M1 A1

(d) Total distance in the first 15 seconds

$S_{15}=\frac{15}{2}\left(2\cdot5+(15-1)\cdot10\right)$

$=\frac{15}{2}(10+140)$

$=\frac{15}{2}(150)=1125$

So the total distance is

$1125\text{ m}=1.125\text{ km}$

$\boxed{1.125\text{ km}}$

📊 Marks: M1 M1 A1

Quick Check

A ball travels $12$ m in the first second and increases by $8$ m each second thereafter.

Find the distance in the $20$ th second and the total distance in the first $20$ seconds.

Solution

$u_{20}=12+19\cdot8=164$

$\boxed{164\text{ m}}$

For the total distance:

$S_{20}=\frac{20}{2}\left(2\cdot12+19\cdot8\right)$

$=10(24+152)=1760$

$\boxed{1760\text{ m}}=1.76\text{ km}$

⚠ Common Mistakes

Using the compound interest formula for a simple interest problem.
Forgetting to write interest rates as decimals in formulas.
Using $t$ instead of $kn$ in compounding frequency questions.
Confusing the $n$ th term formula of an arithmetic sequence with the sum formula.
For distance questions, mixing up “distance in the $n$ th second” and “total distance in the first $n$ seconds”.

📘 Exam Tips

In Paper 1, define the model first: linear growth for simple interest, exponential growth for compound interest.
When comparing two investment models, set the expressions equal before using numerical methods.
For nominal annual rate compounded $k$ times per year, always divide the rate by $k$ and multiply the time by $k$ .
In arithmetic sequence motion problems, identify $u_1$ and $d$ before attempting either $u_n$ or $S_n$ .
Always include units, especially in finance and distance problems.

IB HL Paper 1 Style Exam Questions

Question 1

A sum of $8000 is invested at $4.5\%$ simple interest per annum.

(a) Find the value of the investment after $9$ years.

(b) Another investment of $8000 earns $3.8\%$ compound interest per annum. Find its value after $9$ years.

Question 2

€3500 is invested for $15$ years at a nominal annual rate of $5.4\%$ .

(a) Find the final value if interest is compounded annually.

(b) Find the final value if interest is compounded quarterly.

🧪 HL Challenge Problems

Challenge 1

A savings plan begins with €12 000 invested at $3.2\%$ nominal annual interest, compounded monthly.

(a) Find the value of the investment after $20$ years.

(b) A second account offers simple interest at $4.1\%$ per annum on the same principal. Find its value after $20$ years.

(d) State which model is more realistic for long-term banking and explain why.

Challenge 2

A particle travels distances in successive seconds that form an arithmetic sequence. The first-second distance is $18$ m and the distance increases by $12$ m each second.

(a) Find the distance travelled during the $25$ th second.

(b) Find the total distance travelled in the first $25$ seconds.