What is the difference between a sequence and a series in IB DP Mathematics?

In IB DP Mathematics, a sequence is an ordered list of numbers following a specific pattern, while a series is the sum of the terms of a sequence. For example, in the sequence 2, 4, 6, 8, the series would be the sum 2 + 4 + 6 + 8.

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

To find the nth term of an arithmetic sequence, 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 between consecutive terms. This formula helps in identifying any term in the sequence based on its position.

What is the formula for the sum of an arithmetic series?

The sum of the first n terms of an arithmetic series can be calculated using the formula: S_n = n/2 * (a_1 + a_n), where S_n is the sum of the series, n is the number of terms, a_1 is the first term, and a_n is the nth term. This formula is crucial for solving problems involving the total of a sequence of numbers.

How do you determine if a sequence is geometric?

A sequence is geometric if the ratio between consecutive terms is constant. This constant ratio is called the common ratio, denoted as r. For example, in the sequence 3, 6, 12, 24, the common ratio is 2, as each term is obtained by multiplying the previous term by 2.

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 * (1 - r^n) / (1 - r), where S_n is the sum, a_1 is the first term, r is the common ratio, and n is the number of terms. This formula is applicable when the common ratio r is not equal to 1.

Why is "Applying the AP sum formula to a geometric sequence." a common mistake in Sequence and Series?

Why it happens: Not checking what type of sequence it is. How to avoid it: Compute u_2 - u_1 and u_3 - u_2 — if they match, it's AP. Compute u_2 / u_1 and u_3 / u_2 — if they match, it's GP. NEVER assume.

Why is "Using $S_n = \dfrac{u_1(r^n - 1)}{r - 1}$ when $r = 1$." a common mistake in Sequence and Series?

Why it happens: Formula blindly applied. How to avoid it: When r = 1, every term is u_1, so S_n = n u_1 directly. The GP formula is undefined at r = 1 (division by zero).

Why is "Computing $S_\infty$ for a GP without checking $|r| < 1$." a common mistake in Sequence and Series?

Why it happens: Skipping the qualifying condition. How to avoid it: Always write the line 'Since |r| = < 1, the series converges.' This is an AO3 communication mark.

Why is "On compound-interest problems, rounding $t$ DOWN instead of UP." a common mistake in Sequence and Series?

Why it happens: Habit of rounding to the 'closer' integer. How to avoid it: If the question says 'first exceeds', you need the SMALLEST integer that satisfies the strict inequality. So t > 11.79 means t = 12, NOT 11.

IB DP Mathematics (AA)

Sequence and Series

Q: Why is "Computing $S_\infty$ for a GP without checking $|r| < 1$." a common mistake in Sequence and Series?

Why it happens: Skipping the qualifying condition. How to avoid it: Always write the line 'Since |r| = < 1, the series converges.' This is an AO3 communication mark.

Q: Why is "On compound-interest problems, rounding $t$ DOWN instead of UP." a common mistake in Sequence and Series?

Why it happens: Habit of rounding to the 'closer' integer. How to avoid it: If the question says 'first exceeds', you need the SMALLEST integer that satisfies the strict inequality. So t > 11.79 means t = 12, NOT 11.

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Short Notes - Sequence and Series

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Detailed Study Notes

Detailed notes on Sequences and Series for IB DP Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Sequences and Series — IB Maths AA SL: arithmetic, geometric, sigma notation and applications

AP and GP formulae are the workhorses of Topic 1. This note covers $n$ th-term and sum formulae, sigma notation, infinite geometric series, and the modelling questions (interest, depreciation, growth) that hit every AA SL Paper 2.

At a glance

ARITHMETIC: $u_n = u_1 + (n-1)d$ . Sum: $S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right) = \dfrac{n}{2}(u_1 + u_n)$ .
GEOMETRIC: $u_n = u_1 r^{n-1}$ . Sum: $S_n = \dfrac{u_1(r^n - 1)}{r - 1}$ , $r \ne 1$ .
INFINITE GP: $S_\infty = \dfrac{u_1}{1 - r}$ , valid iff $|r| < 1$ .
SIGMA: $\sum_{k=1}^{n} u_k$ = sum of $u_1$ to $u_n$ .
COMPOUND INTEREST: $A = P\left(1 + \dfrac{r}{100}\right)^t$ (annual) — a GP.
All formulae are IN THE FORMULA BOOKLET. You don't memorise them, but you DO need to use them at speed.

What you’ll learn

Mapped to the IB DP Maths AA SL subject guide (2021 onwards (applies to 2026 exams)).

AO1 — State the AP and GP $n$ th-term and sum formulae.
AO1 — Convert sigma notation to expanded form and back.
AO2 — Solve 'find $n$ ' / 'find $r$ ' / 'find $u_1$ ' problems given multiple terms.
AO2 — Apply infinite GP convergence: identify whether $|r| < 1$ and compute $S_\infty$ .
AO3 — Model and interpret financial / population growth contexts.
AO4 — Use the GDC to compute long sums on Paper 2.

Arithmetic sequences and series

Constant first difference; linear $n$ th term.

An arithmetic sequence has a fixed COMMON DIFFERENCE $d$ : $u_1,\ u_1 + d,\ u_1 + 2d,\ \ldots$

$n$ th term formula (formula booklet): $u_n = u_1 + (n-1)d.$

Sum of the first $n$ terms (formula booklet): $S_n = \frac{n}{2}\Big(2u_1 + (n-1)d\Big) = \frac{n}{2}(u_1 + u_n).$

The second form is faster when you already know $u_n$ .

Worked example. The 4th term of an AP is 11 and the 9th is 26. Find $u_1$ , $d$ and $S_{20}$ .

Set up: $u_4 = u_1 + 3d = 11$ and $u_9 = u_1 + 8d = 26$ .

Subtract: $5d = 15 \Rightarrow d = 3$ , then $u_1 = 2$ .

$S_{20} = \dfrac{20}{2}\big(2(2) + 19(3)\big) = 10(4 + 57) = 610$ .

Why AA SL examiners like this — the two-equation set-up tests AO2 problem solving. State both equations cleanly before subtracting.

$u_n = u_1 + (n-1)d$ .
$S_n = \dfrac{n}{2}(u_1 + u_n)$ is the fastest form when $u_n$ is known.
Subtraction of two indexed terms isolates $d$ .

Geometric sequences and series

Constant ratio; exponential $n$ th term.

A geometric sequence has a fixed COMMON RATIO $r$ : $u_1,\ u_1 r,\ u_1 r^2,\ \ldots$

$n$ th term (formula booklet): $u_n = u_1 r^{n-1}.$

Sum of the first $n$ terms (formula booklet, $r \ne 1$ ): $S_n = \frac{u_1(r^n - 1)}{r - 1}.$

(The form with $r - 1$ is convenient when $r > 1$ . When $r < 1$ , multiply top and bottom by $-1$ to get $\dfrac{u_1(1 - r^n)}{1 - r}$ for nicer numbers.)

Infinite geometric series. If $|r| < 1$ , the partial sums converge: $S_\infty = \frac{u_1}{1 - r}, \quad |r| < 1.$

If $|r| \ge 1$ the series diverges — examiners want you to state this explicitly.

Worked example. A geometric sequence has $u_2 = 6$ and $u_5 = 162$ . Find $u_1$ , $r$ and $S_8$ .

From $\dfrac{u_5}{u_2} = r^3 = \dfrac{162}{6} = 27$ , so $r = 3$ . Then $u_1 = \dfrac{u_2}{r} = 2$ .

$S_8 = \dfrac{2(3^8 - 1)}{3 - 1} = \dfrac{2 \cdot 6560}{2} = 6560$ .

An AP rises in equal steps; a GP doubles each step here. Within six terms the GP pulls dramatically ahead — this is why GP models are used for compound interest, viral spread, and radioactive decay.

$u_n = u_1 r^{n-1}$ .
Ratio of any two terms: $\dfrac{u_{n+k}}{u_n} = r^k$ — solve for $r$ .
Infinite GP converges iff $|r| < 1$ ; otherwise STATE that it diverges.

Sigma notation

Compact way of writing a sum.

$\sum_{k=1}^{n} u_k = u_1 + u_2 + \ldots + u_n.$

The variable $k$ is a dummy index — you can rename it without changing the sum.

Three properties (used constantly in AA SL Paper 1):

$\sum_{k=1}^{n} c \, u_k = c \sum_{k=1}^{n} u_k$ for constant $c$ .
$\sum_{k=1}^{n} (u_k + v_k) = \sum u_k + \sum v_k$ .
$\sum_{k=1}^{n} c = nc$ (constant inside a sum).

Worked example. Evaluate $\sum_{k=1}^{30}(3k - 2)$ .

This is an arithmetic series with $u_1 = 1$ , $d = 3$ , $n = 30$ . $S_{30} = \frac{30}{2}\big(2(1) + 29(3)\big) = 15 \cdot 89 = 1335.$

Worked example (GP via sigma). Evaluate $\sum_{k=1}^{8} 3 \cdot 2^{k-1}$ .

This is a GP with $u_1 = 3$ , $r = 2$ , $n = 8$ : $S_8 = \frac{3(2^8 - 1)}{2 - 1} = 3 \cdot 255 = 765.$

Spotting tip: if the summand contains $k$ LINEARLY (like $3k - 2$ ), it's an AP. If it contains $k$ in an EXPONENT (like $2^{k-1}$ ), it's a GP.

Sigma is linear: split sums and pull constants out.
$\sum_{k=1}^{n} c = nc$ .
Linear in $k$ → AP. Exponential in $k$ → GP.

Modelling applications (Paper 2)

Compound interest, depreciation, population.

Compound interest. $\$ P $invested at$ r%$ p.a. compounded annually grows as $A_t = P\left(1 + \dfrac{r}{100}\right)^t.$

This is a GP with first term $P$ and common ratio $1 + \dfrac{r}{100}$ .

Worked example. $5000 is invested at $4\%$ compound interest per year. After how many years does the amount first exceed $8000?

$5000(1.04)^t > 8000 \Rightarrow 1.04^t > 1.6$ .

Taking $\ln$ : $t > \dfrac{\ln 1.6}{\ln 1.04} \approx \dfrac{0.470}{0.0392} \approx 11.99$ .

So $t = 12$ years (smallest integer satisfying the inequality).

Depreciation. A car worth $25,000 loses $15\%$ of its value each year. The value after $t$ years is $25\,000 \times 0.85^t$ — same structure as compound interest but with $r < 1$ .

Repeated annual payments (annuity). If you deposit $\$ D $each year and the bank pays$ r% $compound, the balance after$ n$ years is the sum of a GP: $\text{Balance} = D \cdot \dfrac{(1+r/100)^n - 1}{r/100}.$

These contexts are AO3 (interpretation) + AO4 (technology). Always:

Define your variables in words.
Set up the AP/GP model.
Compute on the GDC.
State the answer in context with units.

Compound interest is a GP with $r = 1 + \dfrac{\text{rate}}{100}$ .
Depreciation: $r < 1$ — geometric decay.
GDC on Paper 2; manual logs only for non-calculator Paper 1.
ALWAYS interpret the answer in context (AO3).

Quick recap

AP: $u_n = u_1 + (n-1)d$ ; $S_n = \dfrac{n}{2}(u_1 + u_n)$ .
GP: $u_n = u_1 r^{n-1}$ ; $S_n = \dfrac{u_1(r^n - 1)}{r-1}$ , $r \ne 1$ .
$S_\infty = \dfrac{u_1}{1-r}$ iff $|r| < 1$ — otherwise diverges.
Sigma is linear: pull out constants, split sums.
Compound interest = GP with $r = 1 + (\text{rate}/100)$ .

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

$u_n^{AP} = u_1 + (n-1)d$
$S_n^{AP} = \dfrac{n}{2}(u_1 + u_n)$
$u_n^{GP} = u_1 r^{n-1}$
$S_n^{GP} = \dfrac{u_1(r^n - 1)}{r - 1}$
$S_\infty = \dfrac{u_1}{1 - r}$ , $|r| < 1$
Compound interest: $A_t = P(1 + r/100)^t$

How it’s examined

Paper 1: 4–8-mark questions setting up two equations to find $u_1, d$ or $u_1, r$ , then a sum. Paper 2: 10–14-mark modelling questions (compound interest, depreciation, annuity) requiring GDC for arithmetic but algebraic set-up. Examiner reports flag students who don't justify the convergence of an infinite GP — always check $|r|<1$ and state it.

Sources: IB Diploma Programme Mathematics: Analysis and Approaches subject guide (IBO, official). Last reviewed 2026-05-25.

Take this whole topic with you

Step-by-step worked examples — Sequence and Series

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

1Find $d$ and $u_1$ from two terms
Getting started• AP
Question
$u_6 = 23$ and $u_{12} = 47$ in an AP. Find $u_1$ and $d$ .
Step-by-step solution
1. Step 1
  $u_6 = u_1 + 5d = 23$ .
2. Step 2
  $u_{12} = u_1 + 11d = 47$ .
3. Step 3
  Subtract: $6d = 24 \Rightarrow d = 4$ .
4. Step 4
  Substitute: $u_1 + 20 = 23 \Rightarrow u_1 = 3$ .
Answer
$u_1 = 3$ , $d = 4$ .
2Sum of a finite GP
Building confidence• GP
Question
Find $\sum_{k=1}^{10} 5 \cdot 2^{k-1}$ .
Step-by-step solution
1. Step 1
  Identify: $u_1 = 5$ , $r = 2$ , $n = 10$ .
2. Step 2
  $S_{10} = \dfrac{5(2^{10} - 1)}{2 - 1} = 5(1024 - 1) = 5 \cdot 1023 = 5115$ .
Answer
$5115$
3Infinite geometric series
Building confidence• infinite GP
Question
The first term of a GP is 12 and the common ratio is $\dfrac{2}{3}$ . Find $S_\infty$ .
Step-by-step solution
1. Step 1
  Check convergence: $|r| = \tfrac{2}{3} < 1$ ✓ — series converges.
2. Step 2
  $S_\infty = \dfrac{u_1}{1 - r} = \dfrac{12}{1 - 2/3} = \dfrac{12}{1/3} = 36$ .
Answer
$S_\infty = 36$ .
4Compound-interest model (Paper 2)
Stretch• application, AO3
Question
A bank pays $3.5\%$ compound interest per year. $10,000 is deposited. After how many full years does the balance first exceed $15,000?
Step-by-step solution
1. Step 1
  Model: $A_t = 10\,000 \times 1.035^t$ . Solve $10\,000 \times 1.035^t > 15\,000$ .
2. Step 2
  $1.035^t > 1.5 \Rightarrow t > \dfrac{\ln 1.5}{\ln 1.035} \approx \dfrac{0.4055}{0.03440} \approx 11.79$ .
3. Step 3
  Smallest integer $t > 11.79$ is $t = 12$ .
4. Step 4
  Conclusion in context: the balance first exceeds $15,000 after 12 full years.
Answer
12 years.

Key Definitions and Keywords — Sequence and Series

Definitions to memorise and the exact keywords mark schemes credit for sequence and series answers — sharpened from recent examiner reports for the 2026 IB DP Maths AA SL sitting.

Common difference $d$
Examiner keyword
The constant difference between consecutive terms of an AP: $d = u_{n+1} - u_n$ .
Common ratio $r$
Examiner keyword
The constant ratio between consecutive terms of a GP: $r = \dfrac{u_{n+1}}{u_n}$ .
Sigma notation
$\sum_{k=m}^{n} u_k = u_m + u_{m+1} + \ldots + u_n$ . The index $k$ is a dummy variable.
Convergent infinite series
Examiner keyword
A series whose partial sums approach a finite limit. For a GP, this happens iff $|r| < 1$ .

Common Mistakes and Misconceptions — Sequence and Series

The traps other students keep falling into on sequence and series questions — taken from recent IB DP Maths AA SL examiner reports and mark schemes — and how to avoid them.

✕Applying the AP sum formula to a geometric sequence.
Why it happens
Not checking what type of sequence it is.
How to avoid it
Compute $u_2 - u_1$ and $u_3 - u_2$ — if they match, it's AP. Compute $u_2 / u_1$ and $u_3 / u_2$ — if they match, it's GP. NEVER assume.
✕Using $S_n = \dfrac{u_1(r^n - 1)}{r - 1}$ when $r = 1$ .
Why it happens
Formula blindly applied.
How to avoid it
When $r = 1$ , every term is $u_1$ , so $S_n = n u_1$ directly. The GP formula is undefined at $r = 1$ (division by zero).
✕Computing $S_\infty$ for a GP without checking $|r| < 1$ .
Why it happens
Skipping the qualifying condition.
How to avoid it
Always write the line 'Since $|r| = \ldots < 1$ , the series converges.' This is an AO3 communication mark.
✕On compound-interest problems, rounding $t$ DOWN instead of UP.
Why it happens
Habit of rounding to the 'closer' integer.
How to avoid it
If the question says 'first exceeds', you need the SMALLEST integer that satisfies the strict inequality. So $t > 11.79$ means $t = 12$ , NOT 11.