Geometric Sequences & Series — SL 1.3 Geometric sequences and series (finite and infinite) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Long before modern algebra existed, mathematicians noticed that some quantities grow (or shrink) by the same proportion at every step. Ancient Greek scholars studied these patterns in geometry and music, while later thinkers applied them to problems in commerce and astronomy. The idea of a geometric sequence — a list of numbers where each term is obtained by multiplying the previous term by a fixed value — has been a cornerstone of mathematics for over two millennia.

~300 BCE • Euclid's Elements

Euclid defined "geometric proportion" and proved early results about sums of terms in geometric progressions, laying the foundation in Book IX of the Elements.

~250 BCE • Archimedes & the Sand Reckoner

Archimedes used geometric progressions of powers of 10 to estimate the number of grains of sand that could fill the universe, pioneering the idea of exponential scaling.

14th Century • Oresme's Infinite Series

Nicole Oresme, a French mathematician, explored infinite geometric sums and showed that some could converge to a finite value — a breakthrough that would later feed into calculus.

17th–18th Century • Newton, Euler & Formal Series

Isaac Newton and Leonhard Euler developed rigorous formulas for finite and infinite geometric series, applying them to physics, engineering, and number theory.

Modern Era • Finance, Biology & Computing

Today, geometric sequences model compound interest, population growth, radioactive decay, and the algorithms behind digital signal processing.

The central question this topic addresses is straightforward yet powerful: if every term in a sequence is a fixed multiple of the one before it, can we find any term directly and, even more usefully, the sum of many (or infinitely many) terms? The formulas you will learn in this lesson answer both questions elegantly.

SECTION 2

Core Principles & Definitions

A geometric sequence is an ordered list of numbers in which each term after the first is found by multiplying the previous term by a constant called the common ratio, denoted r. The first term is denoted u₁. For example, the sequence 3, 6, 12, 24, … has u₁ = 3 and r = 2. Understanding these building blocks will unlock every formula in this lesson.

Common Ratio (r)

The constant multiplier between consecutive terms. Found by dividing any term by its predecessor: r = u_n+1 ÷ u_n. It can be positive, negative, or a fraction.

General (nth) Term

Any term can be calculated directly without listing all previous terms. The formula u_n = u₁ × rⁿ⁻¹ gives the value at any position.

Finite Geometric Series (Sₙ)

The sum of the first n terms of a geometric sequence. This has a compact closed-form formula that avoids adding terms one by one.

Infinite Geometric Series (S∞)

When |r| < 1, the terms shrink toward zero and the sum approaches a finite limit. This convergent sum is used throughout IB Maths and real-world applications.

✦

KEY TAKEAWAY

Think of a geometric sequence like a chain of photo-copies on a machine that enlarges or shrinks by the same percentage each time. If the machine enlarges by 50 %, every copy is 1.5× the previous one (r = 1.5). If it shrinks by 20 %, every copy is 0.8× the previous one (r = 0.8). The common ratio tells you the scale factor applied at every step.

SECTION 3

Visual Explanation — Geometric Growth

Each bar represents a term in the geometric sequence with u₁ = 2 and r = 2. The pink dashed arrows show that every term is exactly twice the previous one, producing rapid exponential growth.

Notice how the bars do not grow by the same amount — the jump from 2 to 4 is +2, while the jump from 16 to 32 is +16. However, the ratio between consecutive terms is always the same (each term is doubled). This multiplicative pattern, rather than an additive one, is what distinguishes a geometric sequence from an arithmetic one. When r > 1 the terms shoot upward; when 0 < r < 1 the terms decay toward zero; and when r is negative the terms alternate in sign.

SECTION 4

Mathematical Framework

Three key formulas form the backbone of this topic in the IB syllabus. Each appears on the IB formula booklet, so your job is to understand when and how to use them. Below, u₁ is the first term, r is the common ratio, and n is the number of terms.

GENERAL TERM (nTH TERM)

uₙ = u₁ × r ⁿ⁻¹

u_n = the nth term; u₁ = first term; r = common ratio; n = position number. The exponent is n − 1 because the first term uses zero multiplications by r.

SUM OF A FINITE GEOMETRIC SERIES

Sₙ = u₁ × (rⁿ − 1) / (r − 1), r ≠ 1

S_n = the sum of the first n terms. An equivalent form is S_n = u₁ × (1 − rⁿ) / (1 − r), which is often more convenient when |r| < 1.

SUM OF AN INFINITE GEOMETRIC SERIES

S∞ = u₁ / (1 − r), |r| < 1

This formula only works when |r| < 1 (i.e., −1 < r < 1). When this condition holds, the series converges — the partial sums approach a finite limit. If |r| ≥ 1 the series diverges and has no finite sum.

💡

IB Exam Tip

All three formulas are given in the IB formula booklet. In the exam, marks come from choosing the right formula, substituting correctly, and showing clear algebraic steps. Always state the values of u₁, r, and n before substituting.

SECTION 5

Convergence vs. Divergence — A Closer Look

The behaviour of a geometric series depends entirely on the magnitude of the common ratio. When |r| < 1, each successive term becomes smaller and smaller, so the running total settles toward a fixed value — the series converges. When |r| ≥ 1, the terms either stay the same size or grow, so the sum increases without bound — the series diverges.

The green curve (r = ½) and gold curve (r = ⅓) flatten toward their respective S∞ values — they converge. The red curve (r = 2) climbs steeply without bound — it diverges.

Condition on r	Behaviour of Terms	Series Result
\|r\| < 1 (e.g., r = 0.5, −0.3)	Terms shrink toward 0	Converges — use S∞ = u₁ / (1 − r)
\|r\| = 1 (r = 1 or r = −1)	Terms stay same size (or alternate sign)	Diverges — no finite sum
\|r\| > 1 (e.g., r = 2, −3)	Terms grow in magnitude	Diverges — no finite sum

Summary of convergence conditions for an infinite geometric series

SECTION 6

Worked Examples

Example 1 — Finding a Specific Term & Finite Sum

1
Step 1 — Identify the Given Values

A geometric sequence has first term u₁ = 5 and common ratio r = 3. Find the 8th term and the sum of the first 8 terms.
2
Step 2 — Apply the General Term Formula

Using u_n = u₁ × rⁿ⁻¹ with n = 8: u₈ = 5 × 3⁷ = 5 × 2187.

u₈ = 10 935
3
Step 3 — Apply the Finite Sum Formula

S₈ = u₁ × (r⁸ − 1) / (r − 1) = 5 × (3⁸ − 1) / (3 − 1) = 5 × (6561 − 1) / 2 = 5 × 6560 / 2 = 5 × 3280.

S₈ = 16 400

Example 2 — Infinite Sum (Convergent Series)

1
Step 1 — Identify the Given Values

A geometric series has first term u₁ = 12 and common ratio r = −½. Find the sum to infinity.
2
Step 2 — Verify Convergence

|r| = |−½| = ½, which is less than 1. Therefore the series converges and S∞ exists.
3
Step 3 — Substitute into S∞ Formula

S∞ = u₁ / (1 − r) = 12 / (1 − (−½)) = 12 / (1 + ½) = 12 / (3/2) = 12 × 2/3.

S∞ = 8

SECTION 7

Geometric vs. Arithmetic — Key Differences

Students often confuse geometric and arithmetic sequences because both involve a pattern between consecutive terms. The table below highlights the critical differences that will help you choose the correct approach on the IB exam.

Feature	Arithmetic Sequence	Geometric Sequence
Rule between terms	Add a constant (common difference, d)	Multiply by a constant (common ratio, r)
General term	uₙ = u₁ + (n − 1)d	uₙ = u₁ × rⁿ⁻¹
Sum formula	Sₙ = n/2 × (2u₁ + (n − 1)d)	Sₙ = u₁(rⁿ − 1)/(r − 1)
Growth pattern	Linear (constant increase/decrease)	Exponential (accelerating growth or decay)
Infinite sum?	Never converges (unless d = 0)	Converges only when \|r\| < 1
Real-world example	Saving $50 each month	Compound interest at 5 % per year

Side-by-side comparison of arithmetic and geometric sequences

✦

KEY TAKEAWAY

Ask yourself: "Is the sequence growing by a fixed amount (arithmetic) or by a fixed percentage/factor (geometric)?" This single question instantly tells you which set of formulas to use.

SECTION 8

Connections to HL and Further Mathematics

The geometric series framework you have learned at SL level is the gateway to several powerful ideas explored in HL and university-level mathematics. Understanding where these extensions lead can deepen your appreciation of why geometric sequences matter so much.

SL 1.3 Concept	HL / University Extension
Finite geometric sum Sₙ	Partial fraction decomposition and telescoping sums in HL calculus
Infinite sum S∞ when \|r\| < 1	Power series, Taylor/Maclaurin series (HL Calculus), Fourier series (university)
Condition \|r\| < 1 for convergence	Ratio test and root test for general series convergence (university analysis)
Common ratio as a multiplier	Eigenvalues of matrices, geometric growth in differential equations

How SL 1.3 concepts extend into higher-level mathematics

One particularly elegant application is the derivation of the formula for compound interest. When you invest money at a fixed interest rate, your balance each year forms a geometric sequence. Summing regular deposits over time uses the finite geometric series formula directly. In HL, you will also encounter the binomial series, which generalises the idea of an infinite geometric sum to expressions like (1 + x)ⁿ for non-integer n, a tool with wide applications in physics and engineering.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why the infinite geometric series 8 + 4 + 2 + 1 + ½ + … has a finite sum, even though it has infinitely many terms.

Show answer

Each successive term is half the previous one, so the terms shrink rapidly toward zero. Although you keep adding more and more terms, each new addition is so tiny that the running total never exceeds a certain ceiling. Specifically, the common ratio r = ½ satisfies |r| < 1, so the series converges. Using S∞ = u₁ / (1 − r) = 8 / (1 − 0.5) = 16, the sum approaches but never exceeds 16. This is analogous to repeatedly walking half the remaining distance to a wall — you get closer and closer but the total distance covered is finite.

Question 2: BASIC CALCULATION

A geometric sequence has u₁ = 4 and r = 3. Find the 6th term and the sum of the first 6 terms.

Show answer

For the 6th term, use uₙ = u₁ × rⁿ⁻¹: u₆ = 4 × 3⁵ = 4 × 243 = 972. For the sum of the first 6 terms, use Sₙ = u₁(rⁿ − 1)/(r − 1): S₆ = 4 × (3⁶ − 1)/(3 − 1) = 4 × (729 − 1)/2 = 4 × 728/2 = 4 × 364 = 1456. Always verify: the last term (972) should be less than the total sum (1456), which makes sense since earlier terms add up to 484.

Question 3: INTERMEDIATE

The 3rd term of a geometric sequence is 18 and the 6th term is 486. Find the first term, the common ratio, and the sum of the first 10 terms.

Show answer

From uₙ = u₁ × rⁿ⁻¹, set up two equations: u₃ = u₁r² = 18 and u₆ = u₁r⁵ = 486. Dividing the second by the first gives r³ = 486/18 = 27, so r = 3. Substituting back: u₁ × 9 = 18, hence u₁ = 2. Now apply the finite sum formula: S₁₀ = 2 × (3¹⁰ − 1)/(3 − 1) = 2 × (59049 − 1)/2 = 59048. The strategy of dividing two term equations to eliminate u₁ is a standard IB technique worth memorising.

Question 4: APPLIED

A ball is dropped from a height of 10 metres. Each time it bounces, it reaches 60 % of its previous height. Find the total vertical distance travelled by the ball before it comes to rest (assume it bounces infinitely many times).

Show answer

After the initial 10 m drop, the ball rises to 6 m, falls 6 m, rises to 3.6 m, falls 3.6 m, and so on. The total rising distances form a geometric series: 6 + 3.6 + 2.16 + … with u₁ = 6 and r = 0.6. Since |r| < 1 the series converges: S∞ = 6/(1 − 0.6) = 6/0.4 = 15 m. The total falling distances equal 10 + 15 = 25 m (the initial drop plus the same bounces on the way down). Therefore the total distance = 10 + 2 × 15 = 40 metres. This problem illustrates how infinite geometric sums model real-world scenarios involving repeated decay.

Question 5: CRITICAL THINKING

An infinite geometric series has a sum of 20. If the first term is doubled and the common ratio is halved, the new series also sums to 20. Find the original values of u₁ and r.

Show answer

Using S∞ = u₁/(1 − r), the original series gives u₁/(1 − r) = 20, so u₁ = 20(1 − r). For the new series, the first term is 2u₁ and the ratio is r/2, giving 2u₁/(1 − r/2) = 20, so 2u₁ = 20(1 − r/2) = 20 − 10r. Substitute u₁ = 20(1 − r) into 2u₁ = 20 − 10r: 40(1 − r) = 20 − 10r → 40 − 40r = 20 − 10r → 20 = 30r → r = 2/3. Then u₁ = 20(1 − 2/3) = 20 × 1/3 = 20/3. Verify: 20/3 ÷ (1 − 2/3) = 20/3 ÷ 1/3 = 20 ✓. New series: 40/3 ÷ (1 − 1/3) = 40/3 ÷ 2/3 = 20 ✓. This problem rewards careful algebraic manipulation and demonstrates how setting up simultaneous equations from the S∞ formula can determine unknowns.

SUMMARY

Lesson Summary

Summary

A geometric sequence is formed by multiplying each term by a constant called the common ratio (r). The general term formula uₙ = u₁ × rⁿ⁻¹ gives any term directly, while the finite sum Sₙ = u₁(rⁿ − 1)/(r − 1) adds the first n terms efficiently. When |r| < 1, the terms decay toward zero and the infinite sum S∞ = u₁/(1 − r) converges to a finite value.

On the IB exam, always identify u₁ and r first, choose the appropriate formula, and show substitution clearly. Remember: arithmetic means adding a constant, geometric means multiplying by a constant. These sequences appear in real-world contexts from compound interest to radioactive decay, making them one of the most widely applicable tools in the IB syllabus.