# Geometric sequences and series

> IB Mathematics: Applications and Interpretation SL · Number and Algebra
> Source: https://www.owlsprep.com/study/ib-math-ai-sl-u1-geometric-sequences-and-series/

This sub-topic covers core properties of geometric sequences, finite and infinite geometric series, and common applications to real-world growth, decay, and financial problems. These concepts appear regularly in both Paper 1 and Paper 2 IB AI SL exams.

**Prerequisites:** Arithmetic sequences and series; Exponent rules; Basic logarithm operations

## Learning objectives

- Identify geometric sequences and calculate their common ratio
- Find the nth term of any geometric sequence
- Calculate the sum of the first n terms of a finite geometric series
- Determine the sum to infinity of a convergent geometric series
- Solve real-world problems involving geometric growth and decay

## Introduction to Geometric Sequences

**Geometric Sequence** — A sequence where each consecutive term is the product of the previous term and a fixed constant called the common ratio $r$, where $r \neq 0$.

*Notation:* u_n = u_1 r^{n-1}

*Example:* Sequence $2, 6, 18, 54...$ has $u_1 = 2$ and $r=3$

Unlike arithmetic sequences that add a constant difference, geometric sequences multiply by a constant ratio. The common ratio can be positive, negative, or a fraction less than 1.

**Worked example:** Determine if the sequence $4, -2, 1, -0.5, 0.25...$ is geometric. If it is, find the common ratio and the 6th term.

1. Calculate the ratio between each pair of consecutive terms:
2. $$\frac{-2}{4} = -0.5, \quad \frac{1}{-2} = -0.5, \quad \frac{-0.5}{1} = -0.5$$
3. The ratio is constant, so this is a geometric sequence with common ratio $r = -0.5$ and first term $u_1 = 4$.
4. Use the nth term formula to find $u_6$:
5. $$u_6 = u_1 r^{6-1} = 4(-0.5)^5 = 4 \times -\frac{1}{32} = -0.125$$

> **Exam tip:** Always check the ratio between at least three consecutive pairs to confirm a sequence is geometric, don't just check one.

## Finite Geometric Series

**Finite Geometric Series** — The sum of the first $n$ terms of a geometric sequence, calculated with the formula:

*Notation:* S_n

$$S_n = \frac{u_1 (1 - r^n)}{1 - r} = \frac{u_1 (r^n - 1)}{r - 1}, \quad r \neq 1$$

If $r = 1$, the sequence is constant (all terms equal to $u_1$), so the sum simplifies to $S_n = n u_1$. This formula is used for adding up a fixed number of terms, common in annuity and investment problems.

**Worked example:** Find the sum of the first 10 terms of the geometric sequence $3, 6, 12, 24...$

1. Identify the values: $u_1 = 3$, $r = 2$, $n = 10$
2. Substitute into the sum formula:
3. $$S_{10} = \frac{3(2^{10} - 1)}{2 - 1} = 3(1024 - 1) = 3 \times 1023 = 3069$$

## Infinite Geometric Series and Convergence

**Convergent Infinite Geometric Series** — An infinite geometric series approaches a finite sum only when the absolute value of the common ratio is less than 1 ($|r| < 1$). If $|r| \geq 1$, the series diverges and has no finite sum.

*Notation:* S_\infty

*Example:* The series $1 + 0.5 + 0.25 + ...$ converges, $1 + 2 + 4 + ...$ diverges

$$S_\infty = \frac{u_1}{1 - r}, \quad |r| < 1$$

**Worked example:** Find the sum to infinity of the series $12 + 8 + \frac{16}{3} + \frac{32}{9} + ...$

1. Calculate the common ratio:
2. $$r = \frac{8}{12} = \frac{2}{3}$$
3. Check convergence condition: $|r| = \frac{2}{3} < 1$, so the series converges.
4. Substitute into the sum to infinity formula:
5. $$S_\infty = \frac{12}{1 - \frac{2}{3}} = \frac{12}{\frac{1}{3}} = 36$$

> **warning**
>
> Examiners require you to explicitly state the $|r| < 1$ condition before using the sum to infinity formula. You will lose a mark if you skip this step.

## Real-World Applications

Geometric sequences are the standard model for exponential growth and decay in real-world contexts including compound interest, population change, radioactive decay, and recurring investments.

**Worked example:** Find the total value of annual investments of \$1000 made at the start of each year for 10 years, with a constant 5% annual compound interest rate.

1. The first deposit earns interest for 10 years, the second for 9 years, and the last deposit earns interest for 1 year. The total value is a geometric series:
2. $u_1 = 1000 \times 1.05$, $r = 1.05$, $n = 10$
3. Substitute into the finite sum formula:
4. $$S_{10} = \frac{1000 \times 1.05 (1.05^{10} - 1)}{1.05 - 1} \approx 13206.74$$

> **tip**
>
> Always draw a small timeline to confirm the number of compounding periods and first term for application problems to avoid off-by-one errors.

## Common pitfalls

- **Wrong:** Using the nth term formula $u_n = u_1 r^n$ instead of $u_n = u_1 r^{n-1}$
  - Why it fails: This causes an off-by-one exponent error that gives the wrong term, especially for early terms in the sequence
  - Correct: Test the formula with $n=1$: $u_1 = u_1 r^{0} = u_1$, so the exponent must be $n-1$
- **Wrong:** Using the sum to infinity formula without checking $|r| < 1$
  - Why it fails: Examiners explicitly award a mark for this check, and the formula is invalid for divergent series
  - Correct: Always calculate $|r|$ and state that it is less than 1 before applying the $S_\infty$ formula
- **Wrong:** Getting a negative sum for a series with $r>1$
  - Why it fails: Mixing up the order of terms in the numerator of the sum formula
  - Correct: Use $S_n = \frac{u_1(r^n - 1)}{r - 1}$ for $r>1$, which always gives a positive sum matching all positive terms
- **Wrong:** Counting the wrong number of compounding periods for investments
  - Why it fails: Confusing whether the deposit is made at the start or end of the year
  - Correct: Draw a 2-year test case: a deposit at the start of year 1 earns interest for 1 year by the end of year 1, so adjust your exponent accordingly

## Cheatsheet

| Concept | Formula | Required Condition |
| --- | --- | --- |
| nth term | $u_n = u_1 r^{n-1}$ | Always true |
| Sum of first $n$ terms | $S_n = \frac{u_1(1 - r^n)}{1 - r}$ | $r \neq 1$ |
| Sum to infinity | $S_\infty = \frac{u_1}{1 - r}$ | $\|r\| < 1$ |
| Single compound investment | $V_n = P(1 + \frac{R}{100})^n$ | R = annual % rate |
| Annual annuity total | $S_n = \frac{D(1+r)[(1+r)^n - 1]}{r}$ | D = annual deposit |

## What's next

Geometric sequences and series form the foundation for modeling exponential growth and decay, which you will explore further in functions and statistical topics in IB AI SL. This topic is also the basis for financial mathematics applications that make up a significant portion of exam marks, appearing regularly in both Paper 1 and Paper 2. Mastery of convergence and common ratio properties also supports understanding of recursive sequences and probability distributions later in the course.

- [Arithmetic Sequences and Series](https://www.owlsprep.com/study/ib-math-ai-sl-u1-arithmetic-sequences-and-series/)
- [Financial applications of geometric sequences and series](https://www.owlsprep.com/study/ib-math-ai-sl-u1-financial-applications-of-geometric-sequences/)
- [Laws of exponents with rational exponents](https://www.owlsprep.com/study/ib-math-ai-sl-u1-laws-of-exponents-with-rational/)

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