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 2 + 4 + 6 + 8. Understanding the distinction is crucial for solving problems related to sequences and series.

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

To find the nth term of an arithmetic sequence in IB DP Mathematics, 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 preceding terms.

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

The sum of the first n terms of a geometric series in IB DP Mathematics 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.

How can you determine if a series converges or diverges?

In IB DP Mathematics, a series converges if the sum of its terms approaches a finite number as more terms are added. Conversely, it diverges if the sum grows indefinitely. Tests such as the ratio test or the comparison test can be used to determine convergence or divergence of a series.

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

Sequences and series have numerous real-world applications, including calculating interest in finance, analyzing population growth in biology, and solving problems in physics related to motion and waves. Understanding these concepts in IB DP Mathematics helps students apply mathematical reasoning to practical situations.

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

Why it happens: Skipping the qualification. How to avoid it: Always write 'since |r| = < 1, the series converges to S_ = '.

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

Why it happens: Formula blindly applied. How to avoid it: If r = 1, the series is constant, S_n = nu_1.

Why is "Computing $\sum_{k=3}^{n}$ by plugging directly into a $\sum_{k=1}^{n}$ formula." a common mistake in Sequence and Series?

Why it happens: Ignoring the lower limit. How to avoid it: Use \sum_k=3^n u_k = \sum_k=1^n u_k - \sum_k=1^2 u_k.

Sequences and SeriesIB Maths AA HL

Sequence and Series

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

Why it happens: Skipping the qualification. How to avoid it: Always write 'since |r| = < 1, the series converges to S_ = '.

Q: Why is "Using $S_n = u_1(r^n - 1)/(r - 1)$ when $r = 1$." a common mistake in Sequence and Series?

Why it happens: Formula blindly applied. How to avoid it: If r = 1, the series is constant, S_n = nu_1.

Q: Why is "Computing $\sum_{k=3}^{n}$ by plugging directly into a $\sum_{k=1}^{n}$ formula." a common mistake in Sequence and Series?

Why it happens: Ignoring the lower limit. How to avoid it: Use \sum_k=3^n u_k = \sum_k=1^n u_k - \sum_k=1^2 u_k.

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Sequences and Series — IB Maths AA HL: AP/GP, infinite series, sigma sums and the proof-by-induction link

AA HL extends SL sequence work with stronger sigma manipulation, infinite series convergence, and the link to mathematical induction. This note covers all $n$ th-term and sum formulas, infinite GP, financial modelling, and series-sum techniques used in Paper 1 and Paper 3.

At a glance

AP: $u_n = u_1 + (n-1)d$ ; $S_n = (n/2)(2u_1 + (n-1)d) = (n/2)(u_1 + u_n)$ .
GP: $u_n = u_1 r^{n-1}$ ; $S_n = u_1(r^n - 1)/(r - 1)$ , $r \ne 1$ .
INFINITE GP: $S_\infty = u_1/(1 - r)$ , $|r| < 1$ .
SIGMA: $\sum(au_k + v_k) = a\sum u_k + \sum v_k$ ; $\sum c = nc$ ; $\sum k = n(n+1)/2$ ; $\sum k^2 = n(n+1)(2n+1)/6$ .
COMPOUND INTEREST: $A_t = P(1 + r/100)^t$ (discrete GP); $A_t = Pe^{rt/100}$ (continuous).
HL EXTENSION: induction proofs of sum formulas (covered in Proofs subtopic).

What you’ll learn

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

AO1 — Apply AP and GP $n$ th-term and sum formulas.
AO1 — Recognise when a GP converges and compute $S_\infty$ .
AO2 — Solve 'find $u_1$ , $d$ , $r$ ' problems from multiple terms.
AO2 — Use sigma linearity and standard sums.
AO2 — Model financial growth/decay with the appropriate series.
AO3 — Justify convergence with the $|r| < 1$ statement (an explicit mark).

AP and GP — the foundation

All four formulas in the booklet.

Arithmetic sequence with first term $u_1$ and common difference $d$ : $u_n = u_1 + (n - 1)d, \qquad S_n = \frac{n}{2}(2u_1 + (n-1)d) = \frac{n}{2}(u_1 + u_n).$

Geometric sequence with first term $u_1$ and common ratio $r$ : $u_n = u_1 r^{n-1}, \qquad S_n = \frac{u_1(r^n - 1)}{r - 1}, \quad r \ne 1.$

Infinite GP (HL examines convergence rigorously): $S_\infty = \frac{u_1}{1 - r}, \quad |r| < 1.$

If $|r| \ge 1$ , the series DIVERGES — examiners want this stated explicitly.

Worked example (Paper 1). $u_4 = 13$ and $u_{10} = 37$ in an AP. Find $u_1$ , $d$ , $S_{50}$ .

System: $u_1 + 3d = 13$ , $u_1 + 9d = 37$ . Subtract: $6d = 24 \Rightarrow d = 4$ . Then $u_1 = 1$ .

$S_{50} = \dfrac{50}{2}(2 + 49 \cdot 4) = 25 \cdot 198 = 4950$ .

Worked example. A GP has $u_3 = 12$ and $u_6 = 96$ . Find $u_1$ , $r$ , and $S_\infty$ if it converges.

$\dfrac{u_6}{u_3} = r^3 = 8 \Rightarrow r = 2$ . Then $u_1 = u_3 / r^2 = 12/4 = 3$ .

$|r| = 2 \not< 1$ — diverges. State this explicitly.

Worked example. Same data: $u_1 = 27$ , $r = 1/3$ . Find $S_\infty$ .

$|r| = 1/3 < 1$ ✓. $S_\infty = 27 / (1 - 1/3) = 27 \cdot 3/2 = 40.5$ .

Find $r$ from ratio of indexed terms: $u_{n+k}/u_n = r^k$ .
Always check $|r| < 1$ before computing $S_\infty$ .
Two equations from two given terms → solve simultaneously.

Sigma notation and standard sums

Linearity + booklet sums = HL Paper 1 fluency.

Sigma is linear: $\sum_{k=1}^{n}(au_k + v_k) = a \sum_{k=1}^{n} u_k + \sum_{k=1}^{n} v_k, \qquad \sum_{k=1}^{n} c = nc.$

Standard sums (memorise):

$\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$
$\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$ ← booklet
$\sum_{k=1}^{n} k^3 = \left[\dfrac{n(n+1)}{2}\right]^2$

Worked example. Evaluate $\sum_{k=1}^{15} (2k^2 - 3k + 5)$ .

Linearity: $2\sum k^2 - 3\sum k + 5n$ .

$\sum_{k=1}^{15} k^2 = \dfrac{15 \cdot 16 \cdot 31}{6} = 1240$ .

$\sum_{k=1}^{15} k = \dfrac{15 \cdot 16}{2} = 120$ .

Total: $2(1240) - 3(120) + 5(15) = 2480 - 360 + 75 = 2195$ .

Shifting the index. Often the cleanest move: $\sum_{k=3}^{12} u_k = \sum_{k=1}^{12} u_k - \sum_{k=1}^{2} u_k.$

This is essential when the lower limit isn't 1.

Each k adds one bar of height k. The total is n(n+1)/2 — Gauss's famous trick.

Always pull out constants and split sums.
Memorise standard sums up to $\sum k^3$ .
Shift indices by subtracting partial sums when needed.

Financial modelling and Paper 3 connections

Compound interest, annuities, recurring decimals.

Compound interest (discrete, $r$ % annually): $A_t = P\left(1 + \dfrac{r}{100}\right)^t.$

This is a GP with first term $P$ and ratio $1 + r/100$ .

Continuous compounding: $A_t = P e^{rt/100}.$

Worked example. $8000 at $4.5\%$ annual compound interest. How long to triple?

$8000 \cdot 1.045^t = 24000 \Rightarrow 1.045^t = 3 \Rightarrow t = \ln 3 / \ln 1.045 \approx 24.96$ years.

So $t = 25$ years (smallest integer exceeding 24.96).

Annuity (regular deposits). Depositing $D$ at end of each year for $n$ years, $r$ % annual: $\text{Balance} = D \cdot \dfrac{(1 + r/100)^n - 1}{r/100}.$

(Sum of a GP — first term $D$ , ratio $1 + r/100$ , $n$ terms.)

Recurring decimal as a GP. $0.\overline{36} = 0.36 + 0.0036 + \ldots$ is an infinite GP with $u_1 = 0.36$ , $r = 0.01$ .

$S_\infty = 0.36 / 0.99 = 36/99 = 4/11$ .

Worked example (Paper 3 style). A sequence has $u_n = 3n - 2$ for AP terms and $v_n = 2 \cdot 3^{n-1}$ for GP terms. Find $\sum_{n=1}^{10}(u_n + v_n)$ .

$\sum u_n =$ AP sum: $\dfrac{10}{2}(1 + 28) = 145$ .

$\sum v_n =$ GP sum: $\dfrac{2(3^{10} - 1)}{3 - 1} = 3^{10} - 1 = 59048$ .

Total: $145 + 59048 = 59193$ .

Paper 3 routinely combines AP and GP — name each pattern explicitly when setting up.

Compound interest annually = GP; continuously = exponential.
Annuity formula = sum of a GP.
Recurring decimals = infinite GP.
Paper 3: name sequences explicitly when combining.

How it’s examined

Paper 1: exact algebra including infinite GP convergence checks. Paper 2: GDC modelling questions. Paper 3: chained AP/GP, sigma manipulations, links to induction (covered in Proofs). Examiner reports flag students who don't justify the convergence of infinite GP.

Sources: IB Diploma Programme Mathematics: Analysis and Approaches subject guide (IBO, official). Last reviewed 2026-05-31.

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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 $u_1$ and $d$ from two terms
Getting started• AP
Question
$u_5 = 18$ and $u_{15} = 58$ in an AP. Find $u_1, d$ and $S_{50}$ .
Step-by-step solution
1. Step 1
  $u_1 + 4d = 18$ , $u_1 + 14d = 58$ . Subtract: $10d = 40 \Rightarrow d = 4$ .
2. Step 2
  $u_1 = 18 - 16 = 2$ .
3. Step 3
  $S_{50} = (50/2)(2(2) + 49(4)) = 25 \cdot 200 = 5000$ .
Answer
$u_1 = 2$ , $d = 4$ , $S_{50} = 5000$ .
2Infinite GP sum
Building confidence• GP, infinite
Question
$u_1 = 24$ , $r = 5/6$ . Find $S_\infty$ , justifying convergence.
Step-by-step solution
1. Step 1
  $|r| = 5/6 < 1$ ✓ — series converges.
2. Step 2
  $S_\infty = 24/(1 - 5/6) = 24 \cdot 6 = 144$ .
Answer
$S_\infty = 144$ (converges since $|r| < 1$ ).
3Recurring decimal as GP
Stretch• application
Question
Express $0.\overline{72}$ as a fraction in lowest terms.
Step-by-step solution
1. Step 1
  $0.\overline{72} = 0.72 + 0.0072 + \ldots$ — GP with $u_1 = 0.72$ , $r = 0.01$ .
2. Step 2
  $S_\infty = 0.72/0.99 = 72/99 = 8/11$ .
Answer
$8/11$ .
4Sigma — linearity + standard sums
Stretch• sigma
Question
Evaluate $\sum_{k=1}^{12}(4k - k^2)$ .
Step-by-step solution
1. Step 1
  $= 4\sum k - \sum k^2$ .
2. Step 2
  $\sum_{k=1}^{12} k = 78$ , $\sum_{k=1}^{12} k^2 = 12 \cdot 13 \cdot 25/6 = 650$ .
3. Step 3
  Total: $4(78) - 650 = 312 - 650 = -338$ .
Answer
$-338$ .

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 HL sitting.

Common difference $d$
Examiner keyword
Constant gap between consecutive AP terms: $d = u_{n+1} - u_n$ .
Common ratio $r$
Examiner keyword
Constant ratio between consecutive GP terms: $r = u_{n+1}/u_n$ .
Convergent infinite series
Examiner keyword
Partial sums approach a finite limit. For a GP: 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 HL examiner reports and mark schemes — and how to avoid them.

✕Computing $S_\infty$ without checking $|r| < 1$ .
Why it happens
Skipping the qualification.
How to avoid it
Always write 'since $|r| = \ldots < 1$ , the series converges to $S_\infty = \ldots$ '.
✕Using $S_n = u_1(r^n - 1)/(r - 1)$ when $r = 1$ .
Why it happens
Formula blindly applied.
How to avoid it
If $r = 1$ , the series is constant, $S_n = nu_1$ .
✕Computing $\sum_{k=3}^{n}$ by plugging directly into a $\sum_{k=1}^{n}$ formula.
Why it happens
Ignoring the lower limit.
How to avoid it
Use $\sum_{k=3}^{n} u_k = \sum_{k=1}^{n} u_k - \sum_{k=1}^{2} u_k$ .

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