Sequences and series

Q: What is the difference between a sequence and a series?

A sequence is an ordered list of numbers following a rule, whereas a series is the sum of the terms of a sequence.

Q: What does sigma notation (\(\sum\)) represent in mathematics?

Sigma notation represents a compact way of writing the sum of a sequence of terms from a starting index to a finishing index.

Q: What are recurrence relations and how are increasing, decreasing, and periodic sequences classified?

A recurrence relation defines \(u_{n+1}\) in terms of \(u_n\); sequences are increasing if \(u_{n+1} > u_n\), decreasing if \(u_{n+1} < u_n\), and periodic with period \(k\) if \(u_{n+k} = u_n\).

Q: What is an arithmetic progression (AP) and what is the formula for its \(n\)th term?

An AP is a sequence with a constant difference \(d = u_{n+1} - u_n\) between consecutive terms, and its \(n\)th term is given by \(u_n = a + (n - 1)d\).

Q: What are the two formulas for the sum of the first \(n\) terms of an arithmetic progression?

\(S_n = \frac{n}{2}[2a + (n - 1)d]\) and \(S_n = \frac{n}{2}(a + l)\), where \(l\) is the last term.

Q: How is the formula for the sum of an arithmetic series, \(S_n = \frac{n}{2}[2a + (n - 1)d]\), derived?

Write \(S_n\) forwards and backwards, add the two equations term-by-term to get \(2S_n = n(a + l)\), divide by 2, and substitute \(l = a + (n - 1)d\).

Q: What is a geometric progression (GP) and what is the formula for its \(n\)th term?

A GP is a sequence where each term is obtained by multiplying the previous term by a constant ratio \(r = \frac{u_{n+1}}{u_n}\), and its \(n\)th term is \(u_n = a r^{n-1}\).

Q: How is the sum of the first \(n\) terms of a geometric progression, \(S_n = \frac{a(1 - r^n)}{1 - r}\), derived?

Write out \(S_n\), multiply by \(r\) to get \(rS_n\), subtract \(rS_n\) from \(S_n\) to cancel intermediate terms, factorise \(S_n(1 - r) = a(1 - r^n)\), and divide by \((1 - r)\).

Q: Under what condition does an infinite geometric series converge, and what is its sum to infinity \(S_\infty\)?

It converges if and only if \(|r| < 1\) (i.e. \(-1 < r < 1\)), and the sum to infinity is \(S_\infty = \frac{a}{1 - r}\).

Q: How do you solve for an unknown number of terms \(n\) in geometric series inequalities?

Substitute the GP formula into the inequality, isolate the term \(r^n\), take logarithms of both sides, and solve for \(n\) (reversing the inequality if dividing by a negative \(\ln(r)\)).

Mathematics 2210 · CCEA A Level · 9 min read · Updated: 31 Aug 2026

Welcome to Sequences and Series!

Have you ever wondered how banks calculate compound interest on a savings account, or how software engineers model the bouncing of a ball in a video game? The secret behind all of these is sequences and series! Don't worry if maths has felt intimidating in the past — in this guide, we will break down every formula step-by-step with clear real-world analogies, memory tricks, and worked examples designed specifically for your CCEA A2 1 Pure Mathematics exam.

1. The Fundamentals: Sequences, Series, and Sigma Notation

What is a Sequence?

A sequence is simply an ordered list of numbers that follows a particular rule. Each number in the list is called a term.

• We use \(u_1\) to represent the first term, \(u_2\) for the second term, and \(u_n\) for the \(n\)th term (the general term).
• Example: \(2, 5, 8, 11, \dots\) is a sequence where \(u_1 = 2\), \(u_2 = 5\), and so on.

What is a Series?

A series is what you get when you add all the terms of a sequence together.
• Example: \(2 + 5 + 8 + 11 + \dots\)
• We write the sum of the first \(n\) terms as \(S_n\). So for this sequence, \(S_3 = 2 + 5 + 8 = 15\).

Sigma Notation (\(\sum\))

The Greek capital letter Sigma, \(\sum\), means "add them all up". It is shorthand for writing long additions.

\(\sum_{r=1}^{n} u_r = u_1 + u_2 + u_3 + \dots + u_n\)

• The number at the bottom (\(r = 1\)) is the starting value.
• The number at the top (\(n\)) is the finishing value.
• The expression beside it tells you the rule to calculate each term.

Example: Calculate \(\sum_{r=1}^{4} (3r - 1)\):
• When \(r = 1\): \(3(1) - 1 = 2\)
• When \(r = 2\): \(3(2) - 1 = 5\)
• When \(r = 3\): \(3(3) - 1 = 8\)
• When \(r = 4\): \(3(4) - 1 = 11\)
• Sum: \(2 + 5 + 8 + 11 = 26\)

Recurrence Relations (Inductive Definitions)

Sometimes a sequence is defined by telling you how to get the next term from the current term. This is written as a recurrence relation, usually in the form \(u_{n+1} = f(u_n)\).

• Increasing sequence: Each term is strictly greater than the previous one (\(u_{n+1} > u_n\)).
• Decreasing sequence: Each term is strictly less than the previous one (\(u_{n+1} < u_n\)).
• Periodic sequence: The terms repeat in a fixed cycle. If it repeats every \(k\) terms, we say it has period \(k\) (i.e. \(u_{n+k} = u_n\)).

Quick Review: A sequence is a list; a series is a sum. Sigma \(\sum\) means add them up!

2. Arithmetic Progressions (APs)

What is an Arithmetic Progression?

An arithmetic progression (AP) is a sequence where the difference between consecutive terms is always constant. Think of it like walking up a staircase where every step is the exact same height.

• The first term is denoted by \(a\) (or \(u_1\)).
• The constant difference is called the common difference, denoted by \(d\).
• To find \(d\), subtract any term from the term that comes immediately after it: \(d = u_{n+1} - u_n\).

The \(n\)th Term of an AP

Starting at \(a\), you add \(d\) once to get to the 2nd term, you add \(d\) twice to get to the 3rd term, and so on. Therefore, to get to the \(n\)th term, you add \(d\) exactly \((n - 1)\) times:

\(u_n = a + (n - 1)d\)

The Sum of the First \(n\) Terms (\(S_n\))

There are two handy formulas for the sum of an arithmetic series:

1. When you know the first term \(a\) and common difference \(d\):
\(S_n = \frac{n}{2}[2a + (n - 1)d]\)

2. When you know the first term \(a\) and the last term \(l\):
\(S_n = \frac{n}{2}(a + l)\)

Must-Know Proof: Deriving the Formula for \(S_n\)

Did you know? In CCEA A2 exams, you can be asked to prove this formula! Here is the classic method (invented by the famous mathematician Carl Friedrich Gauss as a child):

Step 1: Write out the sum forwards:
\(S_n = a + (a + d) + (a + 2d) + \dots + (l - d) + l\)

Step 2: Write the same sum backwards directly underneath:
\(S_n = l + (l - d) + (l - 2d) + \dots + (a + d) + a\)

Step 3: Add the two equations together vertically term-by-term. Notice that every pair adds up to \((a + l)\):
\(2S_n = (a + l) + (a + l) + (a + l) + \dots + (a + l)\)

Step 4: Since there are \(n\) terms, the right-hand side is simply \(n(a + l)\):
\(2S_n = n(a + l)\)

Step 5: Divide by \(2\):
\(S_n = \frac{n}{2}(a + l)\)

Step 6: Since the last term is \(l = a + (n - 1)d\), substitute this into the formula:
\(S_n = \frac{n}{2}[a + a + (n - 1)d] = \frac{n}{2}[2a + (n - 1)d]\) (Proof complete!)

Worked Example: AP in Action

Question: An arithmetic sequence has a 3rd term of \(11\) and a 9th term of \(35\). Find the sum of the first \(20\) terms.

Solution:
1. Set up equations using \(u_n = a + (n - 1)d\):
\(u_3 = a + 2d = 11\)
\(u_9 = a + 8d = 35\)
2. Subtract the first equation from the second:
\((a + 8d) - (a + 2d) = 35 - 11\)
\(6d = 24 \implies d = 4\)
3. Substitute \(d = 4\) back to find \(a\):
\(a + 2(4) = 11 \implies a + 8 = 11 \implies a = 3\)
4. Calculate \(S_{20}\):
\(S_{20} = \frac{20}{2}[2(3) + (20 - 1)(4)] = 10[6 + 19(4)] = 10[6 + 76] = 10(82) = 820\)

Common Mistake to Avoid: When calculating \(d\), make sure you do \(u_2 - u_1\), not \(u_1 - u_2\). If the sequence is decreasing, \(d\) must be negative!

Key Takeaway: For APs, you add/subtract the same amount each time. Memorise \(u_n = a + (n-1)d\) and practice the proof for \(S_n\)!

3. Geometric Progressions (GPs)

What is a Geometric Progression?

A geometric progression (GP) is a sequence where each term is found by multiplying the previous term by a fixed number. Think of cell division: \(1 \to 2 \to 4 \to 8 \to 16\dots\)

• The first term is denoted by \(a\).
• The multiplier is called the common ratio, denoted by \(r\).
• To find \(r\), divide any term by the preceding term: \(r = \frac{u_{n+1}}{u_n}\).

The \(n\)th Term of a GP

To get from the first term to the \(n\)th term, you multiply by \(r\) a total of \((n - 1)\) times:

\(u_n = a r^{n-1}\)

The Sum of the First \(n\) Terms of a GP

To find the sum \(S_n = u_1 + u_2 + \dots + u_n\), use the formula:

\(S_n = \frac{a(1 - r^n)}{1 - r}\) or \(S_n = \frac{a(r^n - 1)}{r - 1}\) (where \(r \neq 1\))

Tip: Use the first form when \(|r| < 1\) and the second form when \(|r| > 1\) to avoid dealing with double negatives!

Must-Know Proof: Deriving the Formula for GP Sum

This is another high-value exam favourite! Here is the clear step-by-step derivation:

Step 1: Write out the sum \(S_n\):
\(S_n = a + ar + ar^2 + \dots + ar^{n-1}\)

Step 2: Multiply the entire equation by \(r\):
\(rS_n = ar + ar^2 + ar^3 + \dots + ar^n\)

Step 3: Subtract the second equation from the first equation (notice that almost all terms in the middle cancel out!):
\(S_n - rS_n = a - ar^n\)

Step 4: Factorise both sides:
\(S_n(1 - r) = a(1 - r^n)\)

Step 5: Divide by \((1 - r)\) (assuming \(r \neq 1\)):
\(S_n = \frac{a(1 - r^n)}{1 - r}\) (Proof complete!)

Sum to Infinity (\(S_\infty\)) and Convergence

Imagine eating half a pizza, then half of the remainder (\(\frac{1}{4}\)), then half of that (\(\frac{1}{8}\)), and so on forever. You can keep adding pieces forever, but you will never eat more than 1 whole pizza! This is a convergent series.

• An infinite geometric series converges (has a finite total sum) if and only if:

\(|r| < 1\) (which means \(-1 < r < 1\))

• If \(|r| \ge 1\), the sequence diverges, meaning the sum shoots off towards \(\pm \infty\) or oscillates wildly, so no sum to infinity exists.
• When \(|r| < 1\), as \(n \to \infty\), the term \(r^n \to 0\). Substituting \(r^n = 0\) into our sum formula gives the beautiful result:

\(S_\infty = \frac{a}{1 - r}\)

Worked Example: Sum to Infinity

Question: The first term of a GP is \(12\) and its sum to infinity is \(16\). Find the common ratio \(r\) and the sum of the first \(5\) terms.

Solution:
1. Use \(S_\infty = \frac{a}{1 - r}\):
\(16 = \frac{12}{1 - r}\)
\(16(1 - r) = 12\)
\(1 - r = \frac{12}{16} = \frac{3}{4}\)
\(r = 1 - \frac{3}{4} = \frac{1}{4}\)
Notice that \(|\frac{1}{4}| < 1\), so our answer is valid!
2. Find \(S_5\) using \(S_n = \frac{a(1 - r^n)}{1 - r}\):
\(S_5 = \frac{12(1 - (\frac{1}{4})^5)}{1 - \frac{1}{4}} = \frac{12(1 - \frac{1}{1024})}{\frac{3}{4}} = 16\left(1 - \frac{1}{1024}\right) = 16\left(\frac{1023}{1024}\right) = \frac{1023}{64} = 15.984375\)

Key Takeaway: For GPs, you multiply by \(r\) each step. The infinite sum \(S_\infty = \frac{a}{1-r}\) only works when \(|r| < 1\)!

4. Advanced Exam Techniques: Using Logarithms to Find \(n\)

A classic exam question asks: "How many terms are needed for the sum of a GP to exceed a certain value \(K\)?"

Whenever you need to solve for an unknown power \(n\), you must use logarithms.

Step-by-Step Method:

Example: A GP has \(a = 5\) and \(r = 1.2\). Find the smallest value of \(n\) such that \(S_n > 1000\).

Step 1: Set up the inequality using the sum formula:
\(\frac{5(1.2^n - 1)}{1.2 - 1} > 1000\)

Step 2: Simplify the denominator and rearrange:
\(\frac{5(1.2^n - 1)}{0.2} > 1000\)
\(25(1.2^n - 1) > 1000\)
\(1.2^n - 1 > 40\)
\(1.2^n > 41\)

Step 3: Take natural logs (\(\ln\)) or base-10 logs (\(\log_{10}\)) of both sides:
\(\ln(1.2^n) > \ln(41)\)
\(n \ln(1.2) > \ln(41)\)

Step 4: Divide by \(\ln(1.2)\). (Warning: Check if \(\ln(r)\) is negative! Here \(\ln(1.2) \approx +0.1823 > 0\), so the inequality sign does not flip):
\(n > \frac{\ln(41)}{\ln(1.2)}\)
\(n > \frac{3.71357}{0.18232} \approx 20.37\)

Step 5: Choose the smallest integer greater than \(20.37\):
\(n = 21\)

5. Summary and Quick Revision Checklist

Arithmetic Progression (AP) Formula Bank

• \(n\)th term: \(u_n = a + (n - 1)d\)
• Sum of \(n\) terms: \(S_n = \frac{n}{2}[2a + (n - 1)d] = \frac{n}{2}(a + l)\)
• Nature: Linear growth / constant difference \(d\)

Geometric Progression (GP) Formula Bank

• \(n\)th term: \(u_n = a r^{n-1}\)
• Sum of \(n\) terms: \(S_n = \frac{a(1 - r^n)}{1 - r}\)
• Sum to infinity: \(S_\infty = \frac{a}{1 - r}\) (Valid only if \(|r| < 1\))
• Nature: Exponential growth/decay / constant ratio \(r\)

Top Exam Tips

1. Look at the wording: "Added each year" usually means an AP; "percentage increase/decrease each year" usually means a GP.
2. Be careful with \(n\): In word problems, count carefully whether year 1 corresponds to \(n = 1\) or \(n = 0\).
3. Check the convergence condition: If an exam asks "Explain why the series converges", always state that \(|r| < 1\) and show your calculated value of \(r\) lies between \(-1\) and \(1\).
4. Inequality flips: Remember that if \(0 < r < 1\), \(\ln(r)\) is negative. Dividing an inequality by \(\ln(r)\) will reverse the inequality sign!

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the difference between a sequence and a series?

A sequence is an ordered list of numbers following a rule, whereas a series is the sum of the terms of a sequence.

What does sigma notation (\(\sum\)) represent in mathematics?

Sigma notation represents a compact way of writing the sum of a sequence of terms from a starting index to a finishing index.

What are recurrence relations and how are increasing, decreasing, and periodic sequences classified?

A recurrence relation defines \(u_{n+1}\) in terms of \(u_n\); sequences are increasing if \(u_{n+1} > u_n\), decreasing if \(u_{n+1} < u_n\), and periodic with period \(k\) if \(u_{n+k} = u_n\).

What is an arithmetic progression (AP) and what is the formula for its \(n\)th term?

An AP is a sequence with a constant difference \(d = u_{n+1} - u_n\) between consecutive terms, and its \(n\)th term is given by \(u_n = a + (n - 1)d\).

What are the two formulas for the sum of the first \(n\) terms of an arithmetic progression?

\(S_n = \frac{n}{2}[2a + (n - 1)d]\) and \(S_n = \frac{n}{2}(a + l)\), where \(l\) is the last term.

How is the formula for the sum of an arithmetic series, \(S_n = \frac{n}{2}[2a + (n - 1)d]\), derived?

Write \(S_n\) forwards and backwards, add the two equations term-by-term to get \(2S_n = n(a + l)\), divide by 2, and substitute \(l = a + (n - 1)d\).

What is a geometric progression (GP) and what is the formula for its \(n\)th term?

A GP is a sequence where each term is obtained by multiplying the previous term by a constant ratio \(r = \frac{u_{n+1}}{u_n}\), and its \(n\)th term is \(u_n = a r^{n-1}\).

How is the sum of the first \(n\) terms of a geometric progression, \(S_n = \frac{a(1 - r^n)}{1 - r}\), derived?

Write out \(S_n\), multiply by \(r\) to get \(rS_n\), subtract \(rS_n\) from \(S_n\) to cancel intermediate terms, factorise \(S_n(1 - r) = a(1 - r^n)\), and divide by \((1 - r)\).

Under what condition does an infinite geometric series converge, and what is its sum to infinity \(S_\infty\)?

It converges if and only if \(|r| < 1\) (i.e. \(-1 < r < 1\)), and the sum to infinity is \(S_\infty = \frac{a}{1 - r}\).

How do you solve for an unknown number of terms \(n\) in geometric series inequalities?

Substitute the GP formula into the inequality, isolate the term \(r^n\), take logarithms of both sides, and solve for \(n\) (reversing the inequality if dividing by a negative \(\ln(r)\)).

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