# Financial applications of geometric sequences and series > IB Mathematics AI SL · Number and Algebra > Source: https://www.owlsprep.com/study/ib-math-ai-sl-u1-financial-applications-of-geometric-sequences/ This subtopic applies geometric sequences and series to common financial problems including compound interest, asset depreciation, regular savings annuities, and loan repayments. You will learn to model growth and decay for exam questions. **Prerequisites:** [Properties of geometric sequences and series](https://www.owlsprep.com/study/ib-math-ai-sl-u1-geometric-sequences-and-series/) ## Learning objectives - Model compound growth and decay using geometric sequences - Calculate future value of regular savings annuities - Calculate present value of annuities for loans and repayments - Solve for time or interest rate in financial problems using GDC ## Compound Interest and Depreciation **Compound Growth and Decay** — Compound growth (interest) applies a fixed percentage to the current value each period, forming a geometric sequence with common ratio greater than 1. Depreciation is compound decay, with common ratio between 0 and 1. *Example:* A \$1000 investment with 5% annual compound interest forms the sequence 1000, 1050, 1102.5, ... For an initial value $P$, and fixed percentage rate $r$ per period (as a decimal), the value after $n$ periods is given by: $$V_n = P(1+r)^n$$ For depreciation, replace $(1+r)$ with $(1-r)$, so $V_n = P(1-r)^n$. **Worked example:** A car is bought for \$25000 and depreciates by 12% each year. Find its value after 5 years, to the nearest dollar. 1. Identify known values: initial value $P=25000$, annual depreciation rate $r=0.12$, number of years $n=5$. 2. Substitute into the compound decay formula: 3. $$V_5 = 25000(1 - 0.12)^5 = 25000(0.88)^5$$ 4. Calculate the result using a calculator: 5. $$0.88^5 \approx 0.5277, \quad V_5 \approx 25000 \times 0.5277 = 13193$$ > **Exam tip:** Always confirm if the question asks for simple or compound interest. Simple interest is linear, compound is geometric. ## Future Value of an Ordinary Annuity **Ordinary Annuity** — A series of equal regular payments made at the end of each period, earning compound interest. Future value (FV) is the total value of the annuity after all payments are made. The future value of an annuity is the sum of a geometric series, where the first payment earns interest for $n-1$ periods, and the final payment earns no interest. Using the sum of a geometric series, we get the formula: $$FV = C \frac{(1+r)^n - 1}{r}$$ **Worked example:** You deposit \$200 into a savings account at the end of each year, earning 3% annual compound interest. How much will you have after 10 years? 1. Identify values: annual payment $C=200$, rate $r=0.03$, number of payments $n=10$. 2. Substitute into the future value formula: 3. $$FV = 200 \frac{(1+0.03)^{10} - 1}{0.03}$$ 4. Calculate step-by-step: $(1.03)^{10} \approx 1.3439$, so $1.3439 - 1 = 0.3439$ 5. $$FV = 200 \times \frac{0.3439}{0.03} \approx 200 \times 11.463 \approx 2292.60$$ > **Exam tip:** For payments made at the start of each period (annuity due), multiply the result by $(1+r)$ to account for an extra period of interest. ## Present Value of an Annuity for Loans Present value (PV) of an annuity is the current lump-sum value equal to a series of future payments, discounted at the compound interest rate. This is used to calculate the initial value of a loan or the maximum regular withdrawal from a retirement fund. $$PV = C \frac{1 - (1+r)^{-n}}{r}$$ **Worked example:** You take out a 5-year personal loan with 4% annual interest. You make annual repayments of \$1500 at the end of each year. What is the initial loan amount? 1. We need the present value of the annuity: $C=1500$, $r=0.04$, $n=5$. 2. Substitute into the present value formula: 3. $$PV = 1500 \frac{1 - (1.04)^{-5}}{0.04}$$ 4. Calculate: $(1.04)^{-5} \approx 0.8219$, so $1 - 0.8219 = 0.1781$ 5. $$PV = 1500 \times \frac{0.1781}{0.04} \approx 1500 \times 4.4525 \approx 6678.75$$ > **Exam tip:** To find total interest paid on a loan, use: Total Interest = (number of payments × payment amount) - initial loan amount. ## Solving for n or r Exam questions often ask you to find the number of periods $n$ or interest rate $r$ needed to reach a target value. For IB AI SL, you will use your GDC to solve these problems, usually by graphing or trial and error. **Worked example:** How many full years will it take a \$5000 investment to grow to \$8000 at 4% annual compound interest? 1. Set up the compound interest equation: 2. $$5000(1.04)^n = 8000$$ 3. Simplify: divide both sides by 5000 to get $(1.04)^n = 1.6$ 4. Use GDC to graph $y=(1.04)^x - 1.6$ and find the x-intercept, or trial and error: 5. $n=11: (1.04)^{11} \approx 1.5395 < 1.6$; $n=12: (1.04)^{12} \approx 1.6010 > 1.6$ 6. The target is reached after 12 full years. > **Exam tip:** Always round n up to the next whole number of periods, because interest is only added at the end of a period. ## Common pitfalls - **Wrong:** Confusing compound interest with simple interest - Why it fails: Simple interest is linear, compound is geometric, so using the wrong formula gives an incorrect result - Correct: Always check the question wording for 'compound' vs 'simple', remember compound grows faster than simple for the same rate - **Wrong:** Forgetting to adjust for payments at the start of the period - Why it fails: The default ordinary annuity formula assumes payments at the end, so you get a lower value than expected for start-of-period payments - Correct: Multiply the FV or PV result by $(1+r)$ if payments are made at the start of each period - **Wrong:** Rounding intermediate steps too early - Why it fails: Rounding powers or ratios before the final calculation leads to inaccurate results that lose accuracy marks - Correct: Keep all intermediate values on your GDC, only round the final answer to the required precision - **Wrong:** Using $(1+r)$ instead of $(1-r)$ for depreciation - Why it fails: This gives an increasing value instead of a decreasing value, which is wrong for depreciation problems - Correct: For any fixed percentage decrease per period, use $(1 - r)$ as the common ratio - **Wrong:** Rounding n down when solving for time to reach a target - Why it fails: The target is only reached at the end of the period, so rounding down gives a value below the required target - Correct: Always round n up to the next whole number of periods for these problems ## Cheatsheet | Scenario | Formula | | --- | --- | | Compound Interest after n periods | $V_n = P(1+r)^n$ | | Annual Depreciation after n periods | $V_n = P(1-r)^n$ | | Future Value (Ordinary Annuity) | $FV = C \frac{(1+r)^n - 1}{r}$ | | Present Value (Ordinary Annuity) | $PV = C \frac{1 - (1+r)^{-n}}{r}$ | | Total Interest on Loan | $\text{Total Interest} = nC - PV$ | ## What's next This subtopic builds on your understanding of geometric sequences and series to solve practical, frequently tested financial problems for IB AI SL. These concepts appear in both Paper 1 and Paper 2, often as extended response questions worth multiple marks, so fluency with these formulas is critical for exam success. After mastering this content, you can extend your skills to solve more complex problems involving variable interest rates, or use logarithms to find exact solutions for interest rate or time problems. These ideas also connect to exponential modelling which appears later in the syllabus. - [Laws of exponents with rational exponents](https://www.owlsprep.com/study/ib-math-ai-sl-u1-laws-of-exponents-with-rational/) - [Simplification of logarithmic expressions](https://www.owlsprep.com/study/ib-math-ai-sl-u1-simplification-of-logarithmic-expressions/) - [Solving exponential equations using logarithms](https://www.owlsprep.com/study/ib-math-ai-sl-u1-solving-exponential-equations-using-logarithms/) --- From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ib-math-ai-sl-u1-financial-applications-of-geometric-sequences/