IB Mathematics AI SL Financial applications of geometric sequences and series Study Notes - New Syllabus

IB Mathematics AI SL Financial applications of geometric sequences and series Study Notes

LEARNING OBJECTIVE

Financial applications of geometric sequences and series

Key Concepts:

compound interest
annual depreciation.

♦Definition

Simple Interest is the amount paid or earned only on the original principal over a certain time, not on accumulated interest. It is commonly used for short-term loans or investments, typically under one year.

♦Key Terminology

Principal (P): The original amount of money invested or borrowed.
Rate (r): The annual interest rate (expressed as a decimal).
Time (t): Time for which the money is invested or borrowed, in years.
Interest (I): The amount of money earned or paid as interest.

♦General Formula

$I = Prt$

Where:

$I$ = Interest
$P$ = Principal
$r$ = Annual simple interest rate (as a decimal)
$t$ = Time in years

♦Steps to Calculate Simple Interest

1. Convert percentage rate to a decimal:

e.g., $5\% = 0.05$

2. Multiply the decimal rate by the principal:

$P \times r$

3. Multiply by time:

$I = P \times r \times t$

Example

A bank gives $3\%$ simple interest annually.

Tom has £2000 in his account.

How much interest does he earn in 1 year?

▶️Answer/Explanation

Solution:

1. $3\% = 0.03$
2. $0.03 \times 2000 = 60$
Tom earns £60 interest.

Example

For Multiple years

Borrowing £4800 for 3 years at $4\%$ simple interest per year.

What will be the total interest.

▶️Answer/Explanation

Solution:

1. $4\% = 0.04$
2. $0.04 \times 4800 = 192$ per year
3. $192 \times 3 = 576$
Total interest = £576

♦Key Notes

Simple Interest is not compounded.
Over time, simple interest will continue to grow by the same amount each year, while compound interest will grow faster and faster.
It is calculated on the original principal only.
If time is given in months, convert it to years: e.g., 6 months = 0.5 years

♦Compound Interest

Compound interest refers to the process where the interest earned on a principal amount is added back to the principal, and the new amount then earns interest. The formula for compound interest is:

$ FV = PV \left(1 + \frac{r}{100k}\right)^{kn} $

Where:
$FV$ is the future value after $n$ years.
$PV$ is the present value or principal amount.
$r$ is the nominal annual interest rate (as a percentage).
$k$ is the number of compounding periods per year.
$n$ is the number of years.

Example

Suppose we invest \$1000 at an annual interest rate of 6% compounded monthly for 5 years.

Using the compound interest formula, we can find the future value after 5 years.

Using the formula $ FV = PV \left(1 + \frac{r}{100k}\right)^{kn} $

▶️Answer/Explanation

Solution:

$ FV = 1000 \left(1 + \frac{6}{100 \times 12}\right)^{12 \times 5} = 1000 \left(1 + \frac{0.06}{12}\right)^{60} \approx 1349.86 $

Example

Andreas invests $8000$ euros in an account which pays a nominal annual interest rate of $5.25\%$, compounded monthly. Give all answers correct to two decimal places.

Find

(a)the value of the investment after 5 years;

(b) the difference in the final value of the investment if the interest was compounded quarterly at the same nominal rate.

▶️Answer/Explanation

Solution:

a. The value of the investment after 5 years is

$A(t) = P\left(1 + \frac{i}{n}\right)^{nt} = 8000\left(1 + \frac{0.0525}{12}\right)^{12 \times 5} \approx 10395.46$

b. If the interest was compounded quarterly, then the amount would be

$A(t) = P\left(1 + \frac{i}{n}\right)^{nt} = 8000\left(1 + \frac{0.0525}{4}\right)^{4 \times 5} \approx 10383.66$

Difference: $10395.46 – 10383.66$ = €11.80

Solution using GDC

$\text{fx-CG50}$

(a) Go top, scroll to Financial app, l

Press w and type the numbers in followed by leach time (remember that the
present value = investment is entered as a negative number!)

(b)

Interest can be calculated at various compounding frequencies:

Yearly (1 time per year)
Every 6 months (2 times per year)
Every 3 months (4 times per year)
Every month (12 times per year)
Every day (365 times per year)

When interest compounds more frequently than annually, we adjust the formula:

$ FV = PV\left(1 + \frac{r}{100k}\right)^{kn} $

Where:
$ k $ = compounding periods per year
$ n $ = time in years
Other variables remain the same as standard compound interest formula

Example

Suppose you deposit \$2,500 at 4.5% annual interest compounded weekly for 3 years:

Given:
Present Value ($ PV $) = \$2,500
Annual rate ($ r $) = 4.5%
Compounding periods ($ k $) = 52 (weekly)
Time ($ n $) = 3 years

▶️Answer/Explanation

Solution:

$ FV = 2500\left(1 + \frac{4.5}{100 \times 52}\right)^{52 \times 3} $
$ FV = 2500\left(1 + 0.000865\right)^{156} $
$ FV \approx 2500 \times 1.144 $
$ FV \approx \$2,860 $

♦Key Takeaways:
1. More frequent compounding leads to higher returns
2. The formula adjusts by dividing rate and multiplying time by compounding frequency
3. Daily compounding (k = 365) yields slightly more than monthly (k = 12)

♦Depreciation

Depreciation refers to the decrease in value of an asset over time. The formula for depreciation is:

$ V = V_{0} (1 – r)^{t} $

Where:
$V$ is the current value of the asset.
$V_{0}$ is the original value of the asset.
$r$ is the rate of depreciation (as a decimal).
$t$ is the time elapsed.

Example

Using the depreciation formula

Suppose a car is purchased for 20, 000 and depreciates at a rate of 15% per year.

What is the value of the car after 3 years:

▶️Answer/Explanation

Solution:

$V = V_0(1 − r)^t$
$V = 20000(1 − 0.15)^3 ≈ 9261.25$