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Arithmetic & Geometric Progressions

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Cambridge IGCSE · Additional Maths

Arithmetic & Geometric Progressions

Big idea: a sequence is predictable once you know its pattern — arithmetic sequences keep adding the same number, geometric sequences keep multiplying by the same number — and once you know the pattern, you can jump straight to any term or any total without writing the whole list out.

Summary

Everything in this chapter, at a glance

A progression (sequence) is an ordered list of numbers following a rule; a series is what you get when you add those terms up.
Sigma notation (Σ) is shorthand for "add up all these terms" — useful for understanding series, though it won't appear directly in your exam.
In an arithmetic progression, you add the same number (the common difference, d) each time. Formula: u_n = a + (n − 1)d
The sum of an arithmetic series uses S_n = ½n{2a + (n − 1)d} or S_n = ½n{a + l}
In a geometric progression, you multiply by the same number (the common ratio, r) each time. Formula: u_n = ar^{n − 1}
The sum of a geometric series uses S_n = a(1 − rⁿ)/(1 − r), and if |r| < 1, the series converges to a sum to infinity: S_∞ = a/(1 − r)
Questions frequently combine these formulas with simultaneous equations (two unknown terms) or a quadratic that needs solving.

1. The Language of Sequences & Series

Before the formulas — what these words actually mean

What is a progression?

A progression (also called a sequence) is simply an ordered list of numbers that follows a rule. Think of it like a line of dominoes falling — each one knows exactly what comes next, because there's a consistent rule connecting them.

For example: 1, 3, 5, 7, 9, … is a sequence with the rule "start at 1 and add 2 to each number." Every number in this list is called a term.

Rather than writing "the 3rd number in the sequence," mathematicians use subscript notation with the letter u. So for the sequence above:

u₁ = 1 u₂ = 3 u₃ = 5 u₄ = 7 … u_n means "the n-th term" — plug in a term number, get that term's value

Each term can be found by substituting the term number into a formula for the n-th term — this is the whole game of this chapter: find that formula, and you can find any term instantly, without listing everything before it.

What is a series?

A series is what happens when you stop listing the terms and start adding them up. For the sequence 1, 3, 5, 7, … the associated series is 1 + 3 + 5 + 7 + …

We write S_n to mean "the sum of the first n terms":

S_n = u₁ + u₂ + u₃ + … + u_n So for the series above, S₅ = 1 + 3 + 5 + 7 + 9 = 25

Don't mix these up

u_n is one single term (a number sitting in the list). S_n is a running total (everything added together up to that point). This distinction trips people up constantly — a question asking for "the 5th term" wants u₅, not S₅.

Worked Example

Determine the first five terms and the value of S₅ in the progression with terms defined by u_n = 5 − 2n.

Substitute n = 1, 2, 3, 4, 5 one at a time into the formula:
u₁ = 5 − 2(1) = 3
u₂ = 5 − 2(2) = 1
u₃ = 5 − 2(3) = −1
u₄ = 5 − 2(4) = −3
u₅ = 5 − 2(5) = −5

The terms are: 3, 1, −1, −3, −5

To find S₅, add all five terms together:
3 + 1 + (−1) + (−3) + (−5) = −5

S₅ is just the running total — nothing more mysterious than that.

Sigma notation (Σ) — reading it, not testing it

Σ is the Greek capital letter sigma, and in maths it simply means "sum of." The expression to the right tells you what to add up; the numbers above and below the Σ tell you which terms to include.

Example Σ (r = 1 to 5) (2r − 1) = 1 + 3 + 5 + 7 + 9 Substitute r = 1, 2, 3, 4, 5 into (2r − 1) and add the results

The limits don't have to start at 1 — you might see something like Σ (k = 7 to 14) (2k − 13), where you'd substitute k = 7 through 14. The letters r and k are the most commonly used variables here.

Exam note

Sigma notation will not be directly tested in your IGCSE Additional Maths exam — but understanding what it means makes the whole idea of a "series" click much more clearly, which is why it's worth knowing.

2. Arithmetic Progressions

Sequences that keep adding the same amount

What is an arithmetic progression?

In an arithmetic progression, the gap between one term and the next is always exactly the same. That constant gap is called the common difference, d.

Picture a staircase where every step is exactly the same height — that's an arithmetic sequence. Whether you're going up (increasing) or down (decreasing), each step is identical in size.

Example: 2, 5, 8, 11, 14, … first term a = 2 common difference d = +3

Example: 2, 0.5, −1, −2.5, −4, … first term a = 2 common difference d = −1.5

Notice: a positive d means the sequence is climbing (increasing); a negative d means it's falling (decreasing). The size of d tells you how big each step is.

Finding any term: the n-th term formula

The nth term of an arithmetic progression u_n = a + (n − 1)d a = first term · d = common difference · n = which term you want

This formula is given to you on the exam formula sheet — you don't need to memorise how it's derived, just how to use it. Think about why it makes sense though: to get from the 1st term to the n-th term, you take (n − 1) "steps" of size d. That's exactly what a + (n − 1)d calculates.

Two common question types

1. Given one term, find a or d — substitute what you know into the formula and solve the resulting equation.

2. Given two terms, find both a and d — write the formula out for each term to get two equations, then solve them as simultaneous equations.

Worked Example

The first three terms in an arithmetic sequence are 12, k², 5k, … Find two possible values of k.

Since it's arithmetic, there's a single common difference between consecutive terms — so the gap from term 1 to term 2 equals the gap from term 2 to term 3:
d = k² − 12 (2nd term minus 1st term)
d = 5k − k² (3rd term minus 2nd term)

Set the two expressions for d equal to each other:
k² − 12 = 5k − k²
2k² − 5k − 12 = 0

This is now a quadratic — solve by factorising.

(2k + 3)(k − 4) = 0
k = −3/2 or k = 4

Practice Question 1

The 4th term of an arithmetic progression is 17 and the 9th term is 42. Find the first term, a, and the common difference, d.

Practice Question 2

An arithmetic sequence begins 20, 16, 12, … Find the value of the first negative term.

Arithmetic series: adding the terms up

An arithmetic series is what you get when you sum the terms of an arithmetic progression — for the sequence 1, 4, 7, 10, … the series is 1 + 4 + 7 + 10 + …

Sum of the first n terms S_n = ½n{a + l} = S_n = ½n{2a + (n − 1)d} a = first term · l = last term · d = common difference · n = number of terms

You get to choose whichever version is more convenient:

Know the first and last term? Use S_n = ½n{a + l} — it's simpler.
Know the first term and common difference, but not the last term? Use S_n = ½n{2a + (n − 1)d}.

Worked Example

The ninth term of an arithmetic progression is 28. The fifteenth term is 46. The sum of the first n terms is 2500. Show that 3n² + 5n − 5000 = 0, hence find n.

Set up simultaneous equations from the two given terms:
a + 8d = 28 (1)
a + 14d = 46 (2)

(2) − (1): 6d = 18, so d = 3.
Substitute into (1): a + 8(3) = 28, so a = 4.

Substitute a = 4 and d = 3 into the sum formula, set equal to 2500:
n/2 × (2(4) + (n − 1)(3)) = 2500
n(3n + 5) = 5000
3n² + 5n − 5000 = 0 ✓ (shown)

Factorise: (3n + 125)(n − 40) = 0
n = −125/3 or n = 40

n must be a positive integer (you can't have a negative or fractional number of terms), so we reject −125/3.

n = 40

Practice Question 3

Find the sum of the first 20 terms of the arithmetic series 3 + 7 + 11 + 15 + …

3. Geometric Progressions

Sequences that keep multiplying by the same amount

What is a geometric progression?

In a geometric progression, instead of adding a fixed amount each time, you multiply by a fixed amount each time. That multiplier is called the common ratio, r.

Think of it like compound interest, or a population doubling — the change isn't a flat amount, it scales with how big the previous term already was.

Example: 2, 6, 18, 54, 162, … first term a = 2 common ratio r = 3 "start at 2, multiply by 3 each time"

A geometric progression can be increasing (r > 1) or decreasing (0 < r < 1). And if r is negative, something interesting happens — the terms alternate between positive and negative:

Example: 1, −4, 16, −64, 256, … first term a = 1 common ratio r = −4 "start at 1, multiply by −4 each time" — signs flip every term

Arithmetic vs Geometric — the core distinction

Arithmetic: next term = current term + d (constant difference, same "gap" every step)

Geometric: next term = current term × r (constant ratio, same "scale factor" every step)

Finding any term: the n-th term formula

The nth term of a geometric progression u_n = ar^{n − 1} a = first term · r = common ratio · n = which term you want

Why n − 1 and not n? Because to reach the n-th term from the first term, you apply the multiplier (n − 1) times — going from term 1 to term 2 is one multiplication, term 1 to term 3 is two multiplications, and so on.

Finding r when you're given two consecutive terms

Divide a term by the one before it: r = u_n+1 / u_n. Once you have r, substitute it and one known term back into the n-th term formula to find a.

Worked Example

The first three terms in a geometric sequence are 2 − x, 3x, and x² respectively where x < 0. Show that x³ + 7x² = 0. Find the 6th term of the sequence.

Since it's geometric, there's a single common ratio between consecutive terms:
r = 3x / (2 − x) (2nd term ÷ 1st term)
r = x² / 3x (3rd term ÷ 2nd term)

Set the two expressions for r equal:
3x / (2 − x) = x² / 3x
(3x)² = x²(2 − x)
9x² = 2x² − x³
x³ + 7x² = 0 ✓ (shown)

Solve the equation from part (a): x²(x + 7) = 0, so x = 0 or x = −7.

We're told x < 0, so x = −7 (reject x = 0).

Find a and r using x = −7:
a = 2 − (−7) = 9
r = 3(−7) / (2 − (−7)) = −21/9 = −7/3

Substitute into u_n = arⁿ⁻¹ with n = 6:
u₆ = 9 × (−7/3)⁵ = −16 807/27

Practice Question 4

A geometric progression has first term 5 and common ratio 2. Find the value of the first term that exceeds 1000.

Geometric series: adding the terms up

A geometric series is the sum of the terms in a geometric progression.

Sum of the first n terms S_n = a(1 − rⁿ) / (1 − r) or S_n = a(rⁿ − 1) / (r − 1) Use the first version when r < 1, and the second when r > 1 — same formula, just rearranged to avoid negative-over-negative confusion

Both are on the formula sheet — you never need to derive them, only apply them correctly (r cannot equal 1, since you'd be dividing by zero).

Sum to infinity — what happens when the series never ends

Here's something genuinely surprising: if you keep adding infinitely many terms of a geometric series, the total can actually settle on a single, finite number — as long as each term is getting smaller in size.

Think of walking halfway to a wall, then half of the remaining distance, then half of what's left again — you never quite arrive, but the total distance you cover gets closer and closer to a fixed value (the full distance to the wall). That's the intuition behind a sum to infinity.

Sum to infinity — only valid when |r| < 1 S_∞ = a / (1 − r)

Critical condition

This formula only works when |r| < 1 (i.e. −1 < r < 1). If |r| ≥ 1, each term is staying the same size or growing, so the sum keeps climbing forever — it diverges, and no sum to infinity exists. Always check this condition before using the formula!

Worked Example

A geometric progression has first term a. Each term is multiplied by 0.9 to get the next term. The sum of the first 5 terms is 409.51. Find a, and find the sum to infinity of the progression.

Identify what's known: r = 0.9, n = 5, S₅ = 409.51.

Substitute into S_n = a(1 − rⁿ) / (1 − r):
409.51 = a(1 − 0.9⁵) / (1 − 0.9)
409.51 = a(0.40951) / 0.1
409.51 = 4.0951a
a = 100

Since |r| = 0.9 < 1, the sum to infinity exists:
S_∞ = 100 / (1 − 0.9) = 100 / 0.1 = 1000

Practice Question 5

A geometric series has first term 8 and sum to infinity 20. Find the common ratio, r.

What to Memorise

Your formula quick-reference — glance here in the last five minutes before the exam

Concept	Formula	Notes
nth term of an AP	u_n = a + (n − 1)d	a = first term, d = common difference
Sum of an AP (version 1)	S_n = ½n{a + l}	Use when you know first & last term
Sum of an AP (version 2)	S_n = ½n{2a + (n−1)d}	Use when you know a and d, not l
nth term of a GP	u_n = arⁿ⁻¹	a = first term, r = common ratio
Sum of a GP	S_n = a(1−rⁿ)/(1−r)	r ≠ 1; flip to (rⁿ−1)/(r−1) if r > 1
Sum to infinity of a GP	S_∞ = a/(1−r)	Only valid when \|r\| < 1
Finding r (GP)	r = u_n+1 / u_n	Divide any term by the one before it
Finding d (AP)	d = u_n+1 − u_n	Subtract any term from the one after it

Key terms

Progression / Sequence — an ordered list of numbers following a rule.
Series — the sum of the terms of a progression.
Term — a single number in the sequence, written u_n for the n-th one.
Common difference (d) — the fixed amount added each step in an arithmetic progression.
Common ratio (r) — the fixed multiplier applied each step in a geometric progression.
Convergent series — a geometric series where |r| < 1, so it approaches a finite sum to infinity.
Divergent series — a geometric series where |r| ≥ 1, so the sum grows without bound and no sum to infinity exists.

Concepts Checklist

Tick off each idea once you're confident you could explain it out loud

I can explain the difference between a progression and a series
I understand what sigma notation (Σ) means and can read a simple example
I can identify the first term (a) and common difference (d) of an arithmetic sequence
I can use u_n = a + (n − 1)d to find any term, or to find a or d given a term
I can set up and solve simultaneous equations when given two terms of an AP
I can choose the correct sum formula for an arithmetic series and use it confidently
I can identify the first term (a) and common ratio (r) of a geometric sequence
I can use u_n = arⁿ⁻¹ to find any term, or to find a or r given a term
I can find r by dividing consecutive terms, then solve for a
I can use the geometric series sum formula and know when to use each version
I know the condition |r| < 1 must hold before I can use the sum to infinity formula
I can solve a quadratic that arises from setting up an AP or GP problem
I can reject invalid solutions (e.g. negative n, or values outside a given restriction like x < 0)

Exam Tips

What examiners actually look for — and where marks quietly disappear

✓

Both formula pages are given to you. You never need to memorise the n-th term or sum formulas from scratch — but you do need to know which one to reach for and how fast you can locate them. Practise finding them on the formula sheet before the exam so you're not hunting under time pressure.

u_n and S_n are not interchangeable. A huge number of marks are lost by students calculating the wrong one. If the question asks for "the value of the 12th term," that's u₁₂. If it asks for "the total of the first 12 terms," that's S₁₂. Read the question twice.

Forgetting to check |r| < 1 before using S_∞. If r = 1.2 or r = −1.5, there is no sum to infinity — writing down a numerical answer anyway is a common and costly mistake. Always state the check explicitly in your working.

Two given terms means simultaneous equations — every time. Whether it's arithmetic (linear equations) or geometric (often needing division to eliminate a), this is one of the most repeated question structures in this topic. Get very comfortable with the algebra of solving these.

Not rejecting invalid solutions. Quadratics from AP/GP setups often give two solutions, but only one is valid — n must be a positive integer, or a restriction like "x < 0" rules one root out. Examiners specifically award marks for showing you've checked and rejected the invalid answer, not just for finding both roots.

✓

Show your substitution step explicitly. Even when a formula is "given," write out the line where you substitute your values in before simplifying. This is where method marks live, and it also makes it far easier to spot your own arithmetic slips.

Watch your negative common ratios. When r is negative, rⁿ alternates sign depending on whether n is odd or even — a very easy place to make a sign error, especially when raising to odd powers. Double-check with a calculator rather than doing it purely by hand under time pressure.

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