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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.

MAI HL and SL Notes – All topics

Financial Applications

♦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:

\( A = P \left(1 + \frac{r}{n}\right)^{nt} \)

Where:
\(A\) is the total amount of money after \(t\) years.
\(P\) is the principal amount.
\(r\) is the annual interest rate (as a decimal).
\(n\) is the number of times the interest is compounded per year.
\(t\) is the time in 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 total amount of money after 5 years:

▶️Answer/Explanation

Solution:

\( A = P \left(1 + \frac{r}{n}\right)^{nt} = 1000 \left(1 + \frac{0.06}{12}\right)^{(12)(5)} \approx 1349.86 \)

♦Understanding Compound Interest Periods

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:

\( A = P\left(1 + \frac{r}{m}\right)^{mt} \)

Where:
\( m \) = compounding periods per year
\( t \) = 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:
Principal (\( P \)) = $2,500
Annual rate (\( r \)) = 0.045
Compounding periods (\( m \)) = 52 (weekly)
Time (\( t \)) = 3 years

▶️Answer/Explanation

Solution:

\( A = 2500\left(1 + \frac{0.045}{52}\right)^{52 \times 3} \)
\( A = 2500\left(1 + 0.000865\right)^{156} \)
\( A \approx 2500 \times 1.144 \)
\( A \approx \$2,860 \)

♦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 (m=365) yields slightly more than monthly (m=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

Suppose a car is purchased for 20, 000 and depreciates at a rate of 15% per year. Using the depreciation formula, we can find the value of the car after 3 years:

▶️Answer/Explanation

Solution:

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

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