What is the difference between a sequence and a series in Pure Mathematics 2?

In the context of Edexcel International A Levels Pure Mathematics 2, a sequence is an ordered list of numbers following a specific pattern, while a series is the sum of the terms of a sequence. Understanding the distinction is crucial as it forms the basis for further exploration of arithmetic and geometric progressions.

How do you find the nth term of an arithmetic sequence?

To find the nth term of an arithmetic sequence in Pure Mathematics 2, use the formula: a_n = a_1 + (n-1)d, where a_n is the nth term, a_1 is the first term, and d is the common difference. This formula helps in identifying any term in the sequence without listing all previous terms.

What is the formula for the sum of a geometric series?

The sum of the first n terms of a geometric series is given by the formula S_n = a(1 - r^n) / (1 - r), where S_n is the sum, a is the first term, r is the common ratio, and n is the number of terms. This formula is essential for solving problems involving geometric series in Edexcel International A Levels Pure Mathematics 2.

How can you determine if a series converges or diverges?

In Pure Mathematics 2, a series converges if the sum of its terms approaches a finite number as the number of terms increases. Conversely, it diverges if the sum grows indefinitely. For geometric series, convergence occurs when the common ratio |r| < 1. Understanding convergence is key to solving complex series problems in the Edexcel curriculum.

What are some common applications of sequences and series in real-world scenarios?

Sequences and series have numerous applications, such as calculating interest in finance, analyzing population growth in biology, and solving problems in physics and engineering. In the Edexcel International A Levels Pure Mathematics 2, understanding these applications helps students appreciate the practical relevance of mathematical concepts.

Why is "Applying $u_{n} = a + (n - 1) d$ to a geometric sequence" a common mistake in Sequences and series?

Why it happens: Not checking the structure of the sequence before choosing a formula. How to avoid it: Test whether differences are constant (arithmetic) or ratios are constant (geometric). If u_2/u_1 = u_3/u_2 but u_2 - u_1 ≠ u_3 - u_2, it's geometric.

Why is "Using $u_{n} = a + nd$ instead of $u_{n} = a + (n - 1) d$" a common mistake in Sequences and series?

Why it happens: Confusing the index of the term with the number of common differences added. How to avoid it: The first term u_1 has ZERO common differences added: u_1 = a + 0 · d = a. So u_n has (n - 1) differences added. Similarly u_n = a r^n - 1 for geometric.

Why is "Using $S_{\infty} = a/(1 - r)$ when $|r| \geq 1$" a common mistake in Sequences and series?

Why it happens: Plugging into the formula without checking the convergence condition. How to avoid it: ALWAYS check |r| < 1 first. If |r| ≥ 1 the series diverges and S_ does not exist. Examples: r = 2 diverges; r = -1.5 diverges; r = 0.9 converges; r = -0.5 converges. Source: WMA12/01 examiner reports — flagged annually

Why is "Not flipping the inequality when dividing by $\log r$ with $0 < r < 1$" a common mistake in Sequences and series?

Why it happens: Forgetting that of a number < 1 is negative. How to avoid it: If 0 < r < 1 then r < 0. Dividing both sides of an inequality by a NEGATIVE number FLIPS the inequality. Always note the sign of r before dividing.

Why is "Treating '$15\%$ decrease' as multiplication by $0.15$" a common mistake in Sequences and series?

Why it happens: Confusion between the percentage change and the surviving fraction. How to avoid it: A 15\% decrease means 85\% remains, so multiply by 0.85. A 15\% increase means 115\% — multiply by 1.15. The multiplier is 1 - p or 1 + p, never just p.

Pure Mathematics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Sequences and series

Q: What are some common applications of sequences and series in real-world scenarios?

Sequences and series have numerous applications, such as calculating interest in finance, analyzing population growth in biology, and solving problems in physics and engineering. In the Edexcel International A Levels Pure Mathematics 2, understanding these applications helps students appreciate the practical relevance of mathematical concepts.

Q: Why is "Applying $u_{n} = a + (n - 1) d$ to a geometric sequence" a common mistake in Sequences and series?

Why it happens: Not checking the structure of the sequence before choosing a formula. How to avoid it: Test whether differences are constant (arithmetic) or ratios are constant (geometric). If u_2/u_1 = u_3/u_2 but u_2 - u_1 ≠ u_3 - u_2, it's geometric.

Q: Why is "Using $u_{n} = a + nd$ instead of $u_{n} = a + (n - 1) d$" a common mistake in Sequences and series?

Why it happens: Confusing the index of the term with the number of common differences added. How to avoid it: The first term u_1 has ZERO common differences added: u_1 = a + 0 · d = a. So u_n has (n - 1) differences added. Similarly u_n = a r^n - 1 for geometric.

Q: Why is "Using $S_{\infty} = a/(1 - r)$ when $|r| \geq 1$" a common mistake in Sequences and series?

Why it happens: Plugging into the formula without checking the convergence condition. How to avoid it: ALWAYS check |r| < 1 first. If |r| ≥ 1 the series diverges and S_ does not exist. Examples: r = 2 diverges; r = -1.5 diverges; r = 0.9 converges; r = -0.5 converges. Source: WMA12/01 examiner reports — flagged annually

Q: Why is "Not flipping the inequality when dividing by $\log r$ with $0 < r < 1$" a common mistake in Sequences and series?

Why it happens: Forgetting that of a number < 1 is negative. How to avoid it: If 0 < r < 1 then r < 0. Dividing both sides of an inequality by a NEGATIVE number FLIPS the inequality. Always note the sign of r before dividing.

Q: Why is "Treating '$15\%$ decrease' as multiplication by $0.15$" a common mistake in Sequences and series?

Why it happens: Confusion between the percentage change and the surviving fraction. How to avoid it: A 15\% decrease means 85\% remains, so multiply by 0.85. A 15\% increase means 115\% — multiply by 1.15. The multiplier is 1 - p or 1 + p, never just p.

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Sequences and Series Study Notes — Edexcel IAL Mathematics P2 (WMA12/01, 2018 spec — 2026 onwards)

Arithmetic sequences and series, geometric sequences and series, sum to infinity for convergent geometric series, sigma notation, and recurrence relations. The biggest 'modelling' topic in P2 — savings, depreciation, populations.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

P2 5.1 — Generate a sequence from a formula for the $n$ th term or a recurrence relation of the form $u_{n + 1} = f(u_{n})$ .
P2 5.2 — Understand and use sigma notation for sums of series.
P2 5.3 — Recognise arithmetic sequences and series; use the formulae for the $n$ th term and the sum to $n$ terms.
P2 5.4 — Recognise geometric sequences and series; use the formulae for the $n$ th term and the sum to $n$ terms.
P2 5.5 — Know the condition for convergence of an infinite geometric series, and use the sum to infinity.

Arithmetic sequences and series

Constant DIFFERENCE between terms. Use $u_{n} = a + (n - 1)d$ and $S_{n} = \tfrac{n}{2}[2a + (n - 1)d]$ .

Definition. A sequence is arithmetic if every term differs from the previous by a constant, the common difference $d$ . So $u_{n + 1} - u_{n} = d$ .

$n$ th term.

$u_{n} = a + (n - 1) d.$

Note: the first term $u_{1} = a$ — there are $(n - 1)$ common differences added, not $n$ .

Sum to $n$ terms.

$S_{n} = \dfrac{n}{2}[2a + (n - 1) d] = \dfrac{n}{2}(a + l)$

where $l = u_{n}$ is the last term. Both forms are in the IAL formula booklet.

Derivation idea. Write the sum forward and backward, add them: each pair sums to $a + l$ , and there are $n$ pairs (each counted twice). Hence $2 S_{n} = n(a + l)$ .

Worked example. The $3$ rd term is $11$ and the $8$ th term is $26$ . Find $S_{20}$ .

$u_{3} = a + 2d = 11$ ; $u_{8} = a + 7d = 26$ .
Subtract: $5d = 15$ , $d = 3$ . Then $a = 5$ .
$S_{20} = \tfrac{20}{2}[2(5) + 19(3)] = 10 \cdot 67 = 670$ .

Modelling example. Savings: $50 in month 1, increasing by $5 each month. $a = 50$ , $d = 5$ . After $12$ months: $u_{12} = 50 + 11(5) = 105$ (so $105 in month 12). Total over $12$ months: $S_{12} = 6(100 + 55) = 930$ .

Common difference: $d = u_{n + 1} - u_{n}$ .
$u_{n} = a + (n - 1) d$ (note the $-1$ ).
$S_{n}$ : two forms — both in the formula booklet.
Solve simultaneous equations from two known terms.

Geometric sequences and series

Constant RATIO between terms. Use $u_{n} = ar^{n - 1}$ and $S_{n} = \dfrac{a(1 - r^{n})}{1 - r}$ .

Definition. A sequence is geometric if every term is a constant multiple of the previous — the common ratio $r$ . So $u_{n + 1}/u_{n} = r$ .

$n$ th term.

$u_{n} = a r^{n - 1}.$

Note: power is $n - 1$ , not $n$ . The first term has $r^{0} = 1$ .

Sum to $n$ terms.

$S_{n} = \dfrac{a(1 - r^{n})}{1 - r} = \dfrac{a(r^{n} - 1)}{r - 1} \quad (r \neq 1).$

Both forms are in the IAL formula booklet. Use whichever keeps the numerator positive — typically the first for $r < 1$ , the second for $r > 1$ .

Worked example. A geometric series has $a = 24$ , $r = 3/4$ . Find $S_{10}$ .

$S_{10} = \dfrac{24(1 - (3/4)^{10})}{1 - 3/4} = \dfrac{24(1 - 0.0563)}{0.25} = \dfrac{24 \cdot 0.9437}{0.25} \approx 90.60$ .

Sum to infinity. When $|r| < 1$ , the term $r^{n} \to 0$ as $n \to \infty$ , and the series converges to:

$S_{\infty} = \dfrac{a}{1 - r} \quad (|r| < 1).$

(Given in the IAL formula booklet.) When $|r| \geq 1$ the series diverges and no sum to infinity exists.

Worked example. With $a = 24$ , $r = 3/4$ : $S_{\infty} = 24 / (1/4) = 96$ . (And we saw $S_{10} \approx 90.60$ , so the series approaches $96$ rapidly.)

Modelling: depreciation. A car worth $24000 depreciates by $15\%$ each year. After $n$ years: $V_{n} = 24000 \cdot 0.85^{n}$ (multiplier $0.85$ , NOT $0.15$ ). After $5$ years: $\approx \$ 10649 $. Time to fall below \$ 5000: solve $0.85^{n} < 5/24$ , giving $n \geq 10$ .

Common ratio: $r = u_{n + 1} / u_{n}$ .
$u_{n} = a r^{n - 1}$ (power is $n - 1$ ).
$S_{n}$ : two forms, depending on $|r|$ .
$S_{\infty}$ only exists for $|r| < 1$ .

Sigma notation

$\sum_{n = m}^{N} f(n)$ : identify the underlying sequence type, then apply the right sum formula.

Notation.

$\sum_{n = m}^{N} f(n) = f(m) + f(m + 1) + \ldots + f(N).$

The variable $n$ below $\Sigma$ is the index of summation, running from the lower limit $m$ to the upper limit $N$ . Number of terms: $N - m + 1$ .

Strategy. Identify the sequence type from the formula $f(n)$ , then apply the appropriate sum formula.

$f(n)$ has form	Sequence type
$A n + B$ (linear in $n$ )	Arithmetic
$A \cdot b^{n - c}$ or $A b^{n}$	Geometric
$n^{2}, n^{3}, \ldots$	Standard results (not in P2)

Worked example. $\displaystyle \sum_{n = 1}^{20} (3n + 2)$ .

Arithmetic with first term $f(1) = 5$ , last term $f(20) = 62$ , $20$ terms.
$S_{20} = \tfrac{20}{2}(5 + 62) = 670$ .

Worked example. $\displaystyle \sum_{n = 1}^{8} 5 \cdot 2^{n - 1}$ .

Geometric with $a = 5$ , $r = 2$ , $8$ terms.
$S_{8} = \dfrac{5(2^{8} - 1)}{2 - 1} = 5 \cdot 255 = 1275$ .

Calculator note. Most graphing calculators have a $\sum$ button — use it to check your formula-based answer, but always show formula working for method marks.

Limits matter. $\displaystyle \sum_{n = 3}^{10} u_{n} = \sum_{n = 1}^{10} u_{n} - \sum_{n = 1}^{2} u_{n} = S_{10} - S_{2}$ . So if the lower limit isn't $1$ , compute the total minus the early-terms sum.

Number of terms: $N - m + 1$ .
Linear $f(n)$ : arithmetic.
Exponential $f(n)$ : geometric.
Use $S_{N} - S_{m - 1}$ when the lower limit isn't 1.

Recurrence relations

A sequence defined by stating each term in terms of previous ones.

Definition. A recurrence relation defines a sequence using previous terms. Standard form: $u_{n + 1} = f(u_{n})$ , with a stated initial value $u_{1}$ .

Worked example. $u_{n + 1} = 2 u_{n} + 3$ , $u_{1} = 1$ .

$u_{1} = 1$ .
$u_{2} = 2(1) + 3 = 5$ .
$u_{3} = 2(5) + 3 = 13$ .
$u_{4} = 2(13) + 3 = 29$ .
$u_{5} = 2(29) + 3 = 61$ .

Convergent vs divergent recurrences. A recurrence converges to a limit $L$ when $u_{n + 1} \to L$ and $u_{n} \to L$ , so $L = f(L)$ . Solving gives the candidate limit.

Example. $u_{n + 1} = \tfrac{1}{2}(u_{n} + 6)$ . Limit: $L = \tfrac{1}{2}(L + 6) \Rightarrow 2L = L + 6 \Rightarrow L = 6$ . From any starting value the sequence converges to $6$ .

Edexcel-style 'show that' questions. Often a recurrence is paired with an explicit formula and you're asked to verify a specific term, or to prove the limit by induction (the induction direction is rare in P2).

$u_{n + 1} = f(u_{n})$ + initial term defines a sequence.
Limit $L$ (if it exists) satisfies $L = f(L)$ .
Iterate to check or generate specific terms.

How it’s examined

Sequences and series appears in WMA12/01 P2 every sitting (Jan & Jun), typically as Q5-Q11 — worth $10$ - $15$ marks total across multiple questions. Typical questions: find $a$ and $d$ from two terms then a sum (5 marks); modelling problem with arithmetic savings (6-7 marks); geometric series with sum to infinity (5-6 marks); geometric depreciation/growth modelling (6-8 marks); 'find smallest $n$ such that $S_{n} > N$ ' (requires logs — 5 marks); sigma notation evaluation (4-5 marks). Mark schemes apply ECF liberally. A* students should aim for $\geq 90\%$ — modelling questions reward methodical setup and careful interpretation.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA12/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Sequences and series

Step-by-step solutions to past-paper-style questions on sequences and series, written exactly the way a tutor would explain them at the board.

1Arithmetic series — sum to $n$ terms (5 marks, P2)
Core• Adapted from WMA12/01 January 2024 Q6• arithmetic series, sum
Question
The $3$ rd term of an arithmetic sequence is $11$ and the $8$ th term is $26$ . (a) Find the first term $a$ and the common difference $d$ . (b) Find $S_{20}$ , the sum of the first $20$ terms.
Step-by-step solution
1. Step 1
  (a) Use $u_{n} = a + (n - 1) d$ :
  $u_{3} = a + 2d = 11, \qquad u_{8} = a + 7d = 26$
2. Step 2
  Subtract: $5d = 15$ , so $d = 3$ . Then $a = 11 - 2(3) = 5$ .
3. Step 3
  (b) Use $S_{n} = \dfrac{n}{2}[2a + (n - 1) d]$ :
  $S_{20} = \dfrac{20}{2}[2(5) + 19(3)] = 10 \cdot [10 + 57] = 10 \cdot 67 = 670$
Answer
(a) $a = 5$ , $d = 3$ . (b) $S_{20} = 670$ .
Examiner tip
(a) M1 for setting up either equation. A1 for both equations. M1 for solving simultaneously. A1 for $a$ AND $d$ . (b) M1 for using the sum formula. A1 for $670$ . Either sum form ( $S_{n} = \tfrac{n}{2}[2a + (n - 1)d]$ or $S_{n} = \tfrac{n}{2}(a + l)$ where $l$ is the last term) is fully accepted.
2Geometric series — sum to infinity (5 marks, P2)
Core• Adapted from WMA12/01 June 2024 Q5• geometric series, sum to infinity
Question
A geometric series has first term $a = 24$ and common ratio $r = \tfrac{3}{4}$ . (a) Find the sum to infinity $S_{\infty}$ . (b) Find the smallest value of $n$ such that $|S_{\infty} - S_{n}| < 1$ .
Step-by-step solution
1. Step 1
  (a) Since $|r| = 3/4 < 1$ :
  $S_{\infty} = \dfrac{a}{1 - r} = \dfrac{24}{1 - 3/4} = \dfrac{24}{1/4} = 96$
2. Step 2
  (b) $S_{n} = \dfrac{a(1 - r^{n})}{1 - r}$ . So $S_{\infty} - S_{n} = \dfrac{a r^{n}}{1 - r} = 96 r^{n}$ (using $a / (1 - r) = 96$ ).
3. Step 3
  Solve $96 r^{n} < 1$ with $r = 0.75$ :
  $(0.75)^{n} < \dfrac{1}{96}$
4. Step 4
  Take $\log$ (note $\log 0.75$ is negative, so inequality flips):
  $n \log 0.75 < \log(1/96) \;\;\Rightarrow\;\; n > \dfrac{\log(1/96)}{\log 0.75}$
5. Step 5
  Compute: $\log(1/96) = -\log 96 \approx -1.9823$ . $\log 0.75 \approx -0.1249$ . So $n > \dfrac{-1.9823}{-0.1249} \approx 15.87$ .
6. Step 6
  Smallest integer $n$ is $16$ .
Answer
(a) $S_{\infty} = 96$ . (b) $n = 16$ .
Examiner tip
(a) M1 for using $S_{\infty}$ . A1 for $96$ . (b) M1 for the residual $S_{\infty} - S_{n}$ . M1 for the inequality. M1 for taking logs and FLIPPING the sign. A1 for $n = 16$ . The flip is the most common mark-loser: $\log 0.75 < 0$ , so dividing both sides reverses the inequality.
3Arithmetic-series modelling (savings) (6 marks, P2)
Extended• Adapted from WMA12/01 January 2023 Q8• modelling, arithmetic
Question
Anna saves $50 in the first month and increases her saving by $5 each month thereafter. (a) How much does she save in month $12$ ? (b) After how many complete months will her total savings first exceed $5000?
Step-by-step solution
1. Step 1
  Identify: $a = 50$ , $d = 5$ . Month $n$ saving: $u_{n} = 50 + (n - 1) \cdot 5 = 45 + 5n$ .
2. Step 2
  (a) Month $12$ : $u_{12} = 45 + 5(12) = 105$ .
3. Step 3
  (b) Total after $n$ months: $S_{n} = \dfrac{n}{2}[2(50) + (n - 1)(5)] = \dfrac{n}{2}[100 + 5n - 5] = \dfrac{n(95 + 5n)}{2} = \dfrac{5n(n + 19)}{2}$ . Set $S_{n} > 5000$ :
  $\dfrac{5n(n + 19)}{2} > 5000 \;\;\Rightarrow\;\; n(n + 19) > 2000$
4. Step 4
  Solve $n^{2} + 19n - 2000 > 0$ . Quadratic formula: $n = \dfrac{-19 + \sqrt{361 + 8000}}{2} = \dfrac{-19 + \sqrt{8361}}{2} \approx \dfrac{-19 + 91.44}{2} \approx 36.22$ .
5. Step 5
  Smallest integer satisfying the inequality is $n = 37$ .
Answer
(a) $105. (b) $n = 37$ months.
Examiner tip
(a) M1 for using $u_{n}$ . A1 for $105. (b) M1 for using $S_{n}$ . M1 for forming the inequality. M1 for solving the quadratic. A1 for $n = 37$ . Students who use a quadratic equation instead of an inequality but then state the smallest integer typically get full marks.
4Geometric-series modelling (depreciation) (5 marks, P2)
Extended• Adapted from WMA12/01 June 2023 Q9• modelling, geometric, depreciation
Question
A car costs $24000 new. Its value depreciates by $15\%$ each year. (a) Find its value after $5$ years. (b) Find the first year at the END of which the car is worth less than $5000.
Step-by-step solution
1. Step 1
  Model: $V_{n} = 24000 \cdot 0.85^{n}$ where $n$ is years after purchase.
2. Step 2
  (a) $V_{5} = 24000 \cdot 0.85^{5}$ . $0.85^{5} = 0.4437$ , so $V_{5} \approx 24000 \cdot 0.4437 \approx 10648.81$ .
3. Step 3
  (b) Solve $24000 \cdot 0.85^{n} < 5000$ :
  $0.85^{n} < \dfrac{5000}{24000} = \dfrac{5}{24}$
4. Step 4
  Take $\log$ : $n \log 0.85 < \log(5/24)$ . $\log 0.85 \approx -0.0706$ (negative — flip!), so $n > \dfrac{\log(5/24)}{\log 0.85} \approx \dfrac{-0.6812}{-0.0706} \approx 9.65$ .
5. Step 5
  Smallest integer: $n = 10$ .
Answer
(a) $10,649 (nearest dollar). (b) End of year $10$ .
Examiner tip
(a) M1 for the depreciation model. A1 for $10649 (or to nearest cent). (b) M1 for the inequality. M1 for taking logs WITH sign-flip. A1 for $n = 10$ . The 'depreciates by 15%' means multiplied by $0.85$ — not by $0.15$ . A common slip.
5Sigma notation — arithmetic and geometric (5 marks, P2)
Extended• Adapted from WMA12/01 January 2024 Q9• sigma notation
Question
Evaluate (a) $\displaystyle \sum_{n = 1}^{20} (3n + 2)$ and (b) $\displaystyle \sum_{n = 1}^{8} 5 \cdot 2^{n - 1}$ .
Step-by-step solution
1. Step 1
  (a) The terms $3n + 2$ form an arithmetic sequence with first term $3(1) + 2 = 5$ , last term $3(20) + 2 = 62$ , and $20$ terms.
2. Step 2
  Use $S_{n} = \dfrac{n}{2}(a + l) = \dfrac{20}{2}(5 + 62) = 10 \cdot 67 = 670$ .
3. Step 3
  (b) The terms $5 \cdot 2^{n - 1}$ form a geometric sequence with first term $5 \cdot 2^{0} = 5$ , common ratio $2$ , and $8$ terms.
4. Step 4
  Use $S_{n} = \dfrac{a(1 - r^{n})}{1 - r} = \dfrac{5(1 - 2^{8})}{1 - 2} = \dfrac{5(1 - 256)}{-1} = 5 \cdot 255 = 1275$ .
Answer
(a) $670$ . (b) $1275$ .
Examiner tip
(a) M1 for recognising arithmetic structure. A1 for $670$ . (b) M1 for recognising geometric structure. A1 for $1275$ . The trick to sigma notation is to identify the underlying sequence type from the term formula and then apply the appropriate sum formula. Calculators with summation buttons make this trivial — but in 'show that' questions students must show formula work.

Key Formulae — Sequences and series

The formulae you need to memorise for sequences and series on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Arithmetic sequence — $n$ th term
$u_{n} = a + (n - 1) d$
$a$
first term
$d$
common difference
$n$
term number
When to use
Finding any specific term once $a$ and $d$ are known. NOT in the IAL formula booklet — memorise.
Example
$3, 7, 11, 15, \ldots$ has $a = 3$ , $d = 4$ , so $u_{10} = 3 + 9(4) = 39$ .
Arithmetic series — sum to $n$ terms
$S_{n} = \dfrac{n}{2}[2a + (n - 1) d] = \dfrac{n}{2}(a + l)$
$a$
first term
$d$
common difference
$n$
number of terms
$l$
last term, $l = u_{n}$
When to use
Summing the first $n$ terms of an arithmetic sequence. GIVEN in the IAL formula booklet (Pure Mathematics).
Example
First $20$ terms of $5, 8, 11, \ldots$ : $S_{20} = \tfrac{20}{2}[2(5) + 19(3)] = 670$ .
Geometric sequence — $n$ th term
$u_{n} = a r^{n - 1}$
$a$
first term
$r$
common ratio
$n$
term number
When to use
Finding any specific term of a geometric sequence. NOT in the formula booklet — memorise.
Example
$3, 6, 12, 24, \ldots$ has $a = 3$ , $r = 2$ , so $u_{10} = 3 \cdot 2^{9} = 1536$ .
Geometric series — sum to $n$ terms
$S_{n} = \dfrac{a(1 - r^{n})}{1 - r} = \dfrac{a(r^{n} - 1)}{r - 1} \quad (r \neq 1)$
$a$
first term
$r$
common ratio, $r \neq 1$
$n$
number of terms
When to use
Summing the first $n$ terms of a geometric sequence. GIVEN in the IAL formula booklet. Use whichever form keeps the numerator positive.
Example
First $8$ terms of $5, 10, 20, \ldots$ : $S_{8} = \dfrac{5(2^{8} - 1)}{2 - 1} = 1275$ .
Geometric series — sum to infinity
$S_{\infty} = \dfrac{a}{1 - r} \quad (|r| < 1)$
$a$
first term
$r$
common ratio, $|r| < 1$
When to use
When a geometric series converges (i.e. $|r| < 1$ ). GIVEN in the IAL formula booklet. Always check convergence first — if $|r| \geq 1$ , the series diverges and the formula does NOT apply.
Example
$24, 18, 13.5, \ldots$ with $r = 0.75$ gives $S_{\infty} = \dfrac{24}{0.25} = 96$ .

Key Definitions and Keywords — Sequences and series

Definitions to memorise and the exact keywords mark schemes credit for sequences and series answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Sequence
An ordered list of numbers $u_{1}, u_{2}, u_{3}, \ldots$ . Can be defined by a formula for the $n$ th term or by a recurrence relation.
Series
The sum of the terms of a sequence: $S_{n} = u_{1} + u_{2} + \ldots + u_{n}$ . The 'sum to $n$ terms'.
Arithmetic sequence
Examiner keyword
A sequence where each term differs from the previous by a constant — the common difference $d$ . So $u_{n + 1} - u_{n} = d$ .
Example
$5, 8, 11, 14, \ldots$ with $d = 3$ .
Geometric sequence
Examiner keyword
A sequence where each term is a constant multiple of the previous — the common ratio $r$ . So $u_{n + 1} / u_{n} = r$ .
Example
$3, 6, 12, 24, \ldots$ with $r = 2$ .
Convergent (geometric series)
Examiner keyword
A geometric series converges to a finite sum if and only if $|r| < 1$ . In that case the sum to infinity exists and equals $\dfrac{a}{1 - r}$ .
Recurrence relation
A definition of a sequence in terms of one or more previous terms, e.g. $u_{n + 1} = 2 u_{n} + 3$ , $u_{1} = 1$ generates $1, 5, 13, 29, \ldots$ .
Sigma notation
Compact summation notation: $\displaystyle \sum_{r = m}^{n} f(r) = f(m) + f(m + 1) + \ldots + f(n)$ . The variable below $\Sigma$ is the index of summation.
Example
$\displaystyle \sum_{n = 1}^{4} n^{2} = 1 + 4 + 9 + 16 = 30$ .

Common Mistakes and Misconceptions — Sequences and series

The traps other students keep falling into on sequences and series questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Applying $u_{n} = a + (n - 1) d$ to a geometric sequence
Why it happens
Not checking the structure of the sequence before choosing a formula.
How to avoid it
Test whether differences are constant (arithmetic) or ratios are constant (geometric). If $u_{2}/u_{1} = u_{3}/u_{2}$ but $u_{2} - u_{1} \neq u_{3} - u_{2}$ , it's geometric.
✕Using $u_{n} = a + nd$ instead of $u_{n} = a + (n - 1) d$
Why it happens
Confusing the index of the term with the number of common differences added.
How to avoid it
The first term $u_{1}$ has ZERO common differences added: $u_{1} = a + 0 \cdot d = a$ . So $u_{n}$ has $(n - 1)$ differences added. Similarly $u_{n} = a r^{n - 1}$ for geometric.
✕Using $S_{\infty} = a/(1 - r)$ when $|r| \geq 1$
WMA12/01 examiner reports — flagged annually
Why it happens
Plugging into the formula without checking the convergence condition.
How to avoid it
ALWAYS check $|r| < 1$ first. If $|r| \geq 1$ the series diverges and $S_{\infty}$ does not exist. Examples: $r = 2$ diverges; $r = -1.5$ diverges; $r = 0.9$ converges; $r = -0.5$ converges.
✕Not flipping the inequality when dividing by $\log r$ with $0 < r < 1$
Why it happens
Forgetting that $\log$ of a number $< 1$ is negative.
How to avoid it
If $0 < r < 1$ then $\log r < 0$ . Dividing both sides of an inequality by a NEGATIVE number FLIPS the inequality. Always note the sign of $\log r$ before dividing.
✕Treating ' $15\%$ decrease' as multiplication by $0.15$
Why it happens
Confusion between the percentage change and the surviving fraction.
How to avoid it
A $15\%$ decrease means $85\%$ remains, so multiply by $0.85$ . A $15\%$ increase means $115\%$ — multiply by $1.15$ . The multiplier is $1 - p$ or $1 + p$ , never just $p$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Sequences and series — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Sequences and series

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Arithmetic series — sum to nnn terms (5 marks, P2)

2Geometric series — sum to infinity (5 marks, P2)

3Arithmetic-series modelling (savings) (6 marks, P2)

4Geometric-series modelling (depreciation) (5 marks, P2)

5Sigma notation — arithmetic and geometric (5 marks, P2)

Arithmetic sequence — nnnth term

Arithmetic series — sum to nnn terms

Geometric sequence — nnnth term

Geometric series — sum to nnn terms

Geometric series — sum to infinity

Sequence

Series

Arithmetic sequence

Geometric sequence

Convergent (geometric series)

Recurrence relation

Sigma notation

✕Applying un=a+(n−1)du_{n} = a + (n - 1) dun​=a+(n−1)d to a geometric sequence

✕Using un=a+ndu_{n} = a + ndun​=a+nd instead of un=a+(n−1)du_{n} = a + (n - 1) dun​=a+(n−1)d

✕Using S∞=a/(1−r)S_{\infty} = a/(1 - r)S∞​=a/(1−r) when ∣r∣≥1|r| \geq 1∣r∣≥1

✕Not flipping the inequality when dividing by log⁡r\log rlogr with 0<r<10 < r < 10<r<1

✕Treating '15%15\%15% decrease' as multiplication by 0.150.150.15

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Sequences and series — frequently asked questions

1Arithmetic series — sum to $n$ terms (5 marks, P2)

Arithmetic sequence — $n$ th term

Arithmetic series — sum to $n$ terms

Geometric sequence — $n$ th term

Geometric series — sum to $n$ terms

✕Applying $u_{n} = a + (n - 1) d$ to a geometric sequence

✕Using $u_{n} = a + nd$ instead of $u_{n} = a + (n - 1) d$

✕Using $S_{\infty} = a/(1 - r)$ when $|r| \geq 1$

✕Not flipping the inequality when dividing by $\log r$ with $0 < r < 1$

✕Treating ' $15\%$ decrease' as multiplication by $0.15$