How do I tell quickly whether a sequence is arithmetic or geometric?

Test the first few terms both ways. Subtract each term from the next: if you always get the same number, the sequence is arithmetic and that number is $d$. Then divide each term by the one before: if you always get the same number, it is geometric and that number is $r$. A sequence like $2, 6, 10, 14$ has constant differences (arithmetic, $d = 4$); a sequence like $2, 6, 18, 54$ has constant ratios (geometric, $r = 3$). Choosing the wrong type is the most common way to lose every mark on an otherwise easy question.

Why is the exponent $n-1$ and not $n$?

Because the first term has not yet had the step or ratio applied to it. In $u_n = u_1 + (n-1)d$, the 1st term ($n=1$) has taken $0$ steps of $d$, the 2nd term $1$ step, and so on, so the $n$th term has taken $n-1$ steps. The same logic gives $u_n = u_1 r^{n-1}$: the first term has been multiplied by $r$ zero times. Forgetting the $-1$ is the classic off-by-one slip.

How do I turn a percentage increase or decrease into a common ratio $r$?

Work with the multiplier that takes you from 100% to the new percentage. A 7% increase leaves you with 107%, so $r = 1.07$. A 7% decrease leaves you with 93%, so $r = 0.93$. In general, a $p\%$ increase gives $r = 1 + p/100$ and a $p\%$ decrease gives $r = 1 - p/100$. You then multiply by $r$ once for each time period.

A question asks when a term first exceeds some value. How do I solve it on AI SL?

The cleanest AI SL method is the GDC table. Enter the nth-term rule (for example $u_n = 10 \times 1.1^{\,n-1}$), view the table of values, and scroll down to the first integer $n$ whose value passes the target. Write the rule, state the inequality you are solving (such as $10 \times 1.1^{\,n-1} > 100$), then quote the $n$ your table gives — that way the method is on the page even though the GDC found the answer.

When should I reach for the finance solver instead of a series formula?

Use the finance solver (TVM) for compound-interest, loan and annuity questions where money is added or withdrawn regularly, or when you are given a rate and asked for a future balance or a repayment. For a plain 'find the 8th term' or 'sum the first 8 terms' question, the sequence formulas are quicker and make your method visible. When you do use the solver, state the values you entered ($N$, $I\%$, $PV$, $PMT$, $FV$) so the examiner can follow your reasoning.

Arithmetic and geometric sequences and…

Revise Arithmetic and geometric sequences and series for IB Mathematics: Applications and Interpretation (1.2)

Looking for free IB Mathematics: Applications and Interpretation notes on Arithmetic and geometric sequences and series? This Standard Level lesson covers official syllabus point 1.2 on Mathematics: Applications and Interpretation SL. Use the visual steps, key terms, and quick checks below — then practise with IB criterion marking on Arithmetic and geometric sequences and series. Built for students searching "IB Maths AI 1.2", "Arithmetic and geometric sequences and series IB revision", or a free alternative to paid IB note sites.

Arithmetic and geometric sequences and series

From the seats fanning out in a theatre to the balance ticking up in a savings account, huge numbers of real situations are just a starting value plus a rule repeated over and over. Sequences and series are the mathematics of exactly that. This lesson covers the full 1.2 syllabus for AI SL: arithmetic sequences that climb by a fixed step and their sums, geometric sequences that climb (or decay) by a fixed ratio and their sums, sigma notation as shorthand for adding a series up, and the applied contexts — salaries, savings, loans, growth and depreciation — where these models earn their keep. Every AI paper allows a GDC, so the arithmetic is rarely the challenge; the marks are for recognising the model, extracting $u_1$ , $d$ or $r$ and $n$ , and showing the method. As in every calculation topic, the worked examples below are laid out exactly the way our engine marks your own practice.

BY THE END, YOU CAN…

1Distinguish arithmetic sequences (constant difference $d$ ) from geometric sequences (constant ratio $r$ ), and find $u_1$ , $d$ and $r$ from the terms of a sequence.
2Use $u_n = u_1 + (n-1)d$ and $S_n = \frac{n}{2}(2u_1+(n-1)d)$ for arithmetic sequences and series, and $u_n = u_1 r^{n-1}$ and $S_n = \frac{u_1(r^n-1)}{r-1}$ for geometric sequences and series.
3Read, expand and evaluate a finite series written in sigma notation, and convert between a series and its sigma form.
4Model applied situations — salaries, savings plans, loan balances, growth and depreciation — as arithmetic or geometric, choosing the right formula and interpreting the answer in context.
5Use a GDC efficiently (nth-term rules, tables and the finance solver) while still presenting the method so that method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Two Ways a Pattern Can Grow

Almost every pattern in this topic grows in one of two ways: by adding the same amount each step, or by multiplying by the same amount each step. Spot which one you are looking at, pull out the starting value and the step (or the ratio), and a single formula from the booklet does the rest — for one term or for a whole running total.

THINK OF IT LIKE…

Picture two savers. The first tucks exactly $50 under the mattress every month: the pile climbs in equal steps — that is arithmetic. The second puts money in an account that grows by 3% each month, so each month's growth is bigger than the last: the balance climbs by multiplication — that is geometric. Adding a fixed amount is a straight-line climb; multiplying by a fixed factor is a curve that bends upward (or, if the factor is less than one, decays towards zero).

1
Decide the type. Look at consecutive terms. If you SUBTRACT and always get the same number, it is arithmetic with common difference $d$ . If you DIVIDE and always get the same number, it is geometric with common ratio $r$ .
2
Pull out the parameters. Write down the first term $u_1$ , the step $d$ or ratio $r$ , and the term number $n$ the question asks about. Read percentage changes carefully: a 4% rise means $r = 1.04$ , a 4% fall means $r = 0.96$ .
3
Pick the formula. One term uses the $u_n$ formula; a running total uses the $S_n$ formula. All four are in your formula booklet — the skill is choosing, not memorising.
4
Substitute, compute, interpret. Let the GDC do the arithmetic, then round sensibly and check the answer makes sense for the context — money to 2 decimal places, people to a whole number.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

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

Full topic notes

Formal explanation with the rigour you need for the exam.

Arithmetic sequences: adding a constant step

An arithmetic sequence adds (or subtracts) the same amount to move from one term to the next. That fixed amount is the common difference $d$ , and you find it by subtracting any term from the one after it: $d = u_{n+1} - u_n$ . If $d > 0$ the sequence rises steadily; if $d < 0$ it falls. Because the step never changes, an arithmetic sequence plotted against term number lies on a straight line — its terms are a linear pattern.

$n$ th term of an arithmetic sequence: $u_n = u_1 + (n-1)d$ , \n where $u_1$ is the first term, $n$ is the term number, and $d$ is the common difference. The step is applied $n-1$ times to reach the $n$ th term.

Adding the terms of an arithmetic sequence gives an arithmetic series. Rather than adding term by term, use the sum formula. There are two equivalent versions: reach for the first when you know the difference, and the second when you already know the last term $u_n$ .

Sum of the first $n$ terms: $S_n = \dfrac{n}{2}\bigl(2u_1 + (n-1)d\bigr)$ , \n or, when the last term $u_n$ is known, $S_n = \dfrac{n}{2}(u_1 + u_n)$ .

Common difference $d$ : subtract consecutive terms, $d = u_{n+1} - u_n$ . A constant difference is what makes a sequence arithmetic.
nth term: $u_n = u_1 + (n-1)d$ — one term of the sequence.
Sum: $S_n = \frac{n}{2}(2u_1 + (n-1)d)$ — the running total of a series. Use $\frac{n}{2}(u_1 + u_n)$ if the last term is known.

Geometric sequences: multiplying by a constant ratio

A geometric sequence multiplies by the same amount to move from one term to the next. That fixed multiplier is the common ratio $r$ , found by dividing any term by the one before it: $r = \frac{u_{n+1}}{u_n}$ . This is the crucial difference from arithmetic — you DIVIDE to find $r$ , you do not subtract. When $|r| > 1$ the terms grow ever faster; when $0 < |r| < 1$ they decay towards zero, which is exactly how depreciation works.

$n$ th term of a geometric sequence: $u_n = u_1 r^{\,n-1}$ , \n where $u_1$ is the first term, $n$ is the term number, and $r$ is the common ratio. The first term is multiplied by $r$ a total of $n-1$ times.

Applied questions usually hand you $r$ as a percentage change. Convert it to a multiplier: a growth of $p\%$ gives $r = 1 + p/100$ , and a decay of $p\%$ gives $r = 1 - p/100$ . So a population growing 3% a year has $r = 1.03$ ; a car losing 18% a year retains 82% of its value, so $r = 0.82$ .

Sum of a geometric series

Adding the terms of a geometric sequence gives a geometric series, which models the total of a repeatedly-scaling quantity — money paid into a growing savings plan, or the total distance a bouncing ball travels. Because each term is bigger (or smaller) than the last by the factor $r$ , the sum has its own compact formula.

Sum of the first $n$ terms of a geometric series: \n $S_n = \dfrac{u_1(r^n - 1)}{r - 1} = \dfrac{u_1(1 - r^n)}{1 - r}, \quad r \neq 1.$ \n The two forms are equal; pick the one that keeps the denominator positive to avoid a sign slip.

Sigma notation

Sigma notation is compact shorthand for a series. The Greek capital sigma $\sum$ means 'add up', the letter and number below give where the counter starts, and the number above gives where it stops. So $\sum_{k=1}^{4}(2k+1)$ says: substitute $k = 1, 2, 3, 4$ into $2k+1$ and add the results — that is $3 + 5 + 7 + 9 = 24$ . You will meet arithmetic and geometric series wrapped in this notation, and the skill is simply to expand a few terms, decide the type, and reach for the matching sum formula.

$\displaystyle \sum_{k=1}^{n} u_k = u_1 + u_2 + \cdots + u_n.$ \n The counter $k$ runs from the lower value to the upper value; each value is substituted into $u_k$ and the results are added.

Common mistakes examiners penalise

Mixing up arithmetic and geometric — always test consecutive terms both ways first. Constant DIFFERENCE means arithmetic ( $d$ ); constant RATIO means geometric ( $r$ ). Using the wrong family of formula loses the whole question.
The off-by-one on $n$ — the exponent and bracket are $n-1$ , not $n$ . The 10th term uses $(10-1) = 9$ steps. And 'after $n$ years' from a starting value is often the $(n+1)$ th term, because the starting value is the 1st term.
Finding $r$ by subtracting — the common ratio is $u_{n+1} \div u_n$ , not $u_{n+1} - u_n$ . Dividing consecutive terms is the only correct way; subtracting gives the difference, which is meaningless for a geometric sequence.
Confusing a term with a total — 'find the 8th term' needs the $u_n$ formula; 'find the total of the first 8 terms' needs the $S_n$ formula. Read whether the question wants one value or a running sum.
Wrong ratio from a percentage — a 15% decrease is $r = 0.85$ , not $0.15$ or $-0.15$ ; a 15% increase is $r = 1.15$ . Convert the percentage to the multiplier that takes 100% to the new value.
Over-rounding mid-calculation — carry the GDC's full figures and round only the final answer, sensibly for the context (money to 2 d.p., people to a whole number, otherwise 3 s.f.).
Writing no method with a GDC answer — a GDC is allowed on every AI paper, but a lone number earns no method mark. Show the formula, the substitution, or the finance-solver inputs so the working is on the page.

Model answer — marked the way our engine marks it

IB marks are analytic: each mark is tied to a specific line of working. A method mark (M) rewards a correct approach — quoting a formula, substituting correctly, setting up the finance solver — even if a later number is wrong. An accuracy mark (A) rewards a correct value, and it DEPENDS on the method mark it follows: no method, no accuracy. Follow-through (FT) means an early slip need not cost the marks that build on it, as long as the later step is done correctly on your own figures. A GDC is allowed on every AI paper, so the accuracy is expected to be exact — but that protection only exists if the method is written down. Study how each of the 5 marks below is earned by a specific line.

Where this leads

Sequences and series are the backbone of the whole 'Number and algebra' unit. The geometric model you built here powers compound interest, depreciation, loan and annuity calculations, where the same $r^n$ growth is dressed up in financial language and solved with the finance solver. The habit of reading a percentage change as a multiplier reappears throughout financial mathematics, and the discipline of naming $u_1$ , $d$ or $r$ and $n$ before touching a formula is exactly what every applied modelling question rewards. Master the two ways a pattern can grow — add a step, or multiply by a ratio — and a large slice of AI SL becomes variations on a pattern you already know how to describe.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A graduate starts a job on an annual salary of 28 000 dollars. Each year the salary rises by 750 dollars. \n (a) Find the salary in the 10th year. \n (b) Find the total salary earned over the first 10 years. [5]

1
The salary rises by a fixed amount each year, so this is arithmetic with $u_1 = 28000$ and $d = 750$ .\n\n**(a) Salary in the 10th year ( $n = 10$ ).\nUse the nth-term formula:\n $u_{10} = u_1 + (10-1)d = 28000 + 9 \times 750$ [M1: nth-term formula with $n-1 = 9$ ]\n $u_{10} = 28000 + 6750 = 34750.$ \nThe salary in the 10th year is 34 750 dollars. [A1]\n\n(b) Total over the first 10 years.**\nThis is a sum, so use the series formula:\n $S_{10} = \dfrac{10}{2}\bigl(2 \times 28000 + (10-1) \times 750\bigr)$ [M1: sum formula]\n $S_{10} = 5\,(56000 + 6750) = 5 \times 62750 = 313750.$ [A1]\nThe total earned over the first 10 years is 313 750 dollars. [A1]\n\nNotice part (b) uses the SUM formula, not the term formula — a total always calls for $S_n$ .

✓ FULLY WORKED

EXAMPLE 2

A car is bought for 25 000 dollars and depreciates by 18% each year. \n Find its value, to the nearest dollar, after 5 years. [4]

1
The value is multiplied by the same factor each year, so this is geometric.\n\nIdentify the parameters. $u_1 = 25000$ . Losing 18% leaves 82%, so $r = 0.82$ . 'After 5 years', with the purchase price as the 1st term, is the 6th term, so $n = 6$ . [M1: correct $r$ and $n$ ]\n\nApply the nth-term formula.\n $u_6 = u_1 r^{\,6-1} = 25000 \times 0.82^{5}$ [M1: geometric nth-term formula]\n $u_6 = 25000 \times 0.3707\ldots = 9265.58\ldots$ [A1]\n\nRound for the context. Money to the nearest dollar:\n $u_6 \approx 9266$ dollars. [A1]\n\nThe ' $0.82$ ' (not $-0.18$ ) captures the decay, and the ' $n = 6$ ' reflects that the initial value is the first term.

✓ FULLY WORKED

EXAMPLE 3

A saver deposits 20 dollars in the first month and increases each monthly deposit by 5%. \n Find the total amount deposited after 2 years, to the nearest cent. [4]

1
Each deposit is 5% larger than the last, so the deposits form a geometric sequence and their total is a geometric series.\n\nIdentify the parameters. $u_1 = 20$ ; a 5% increase gives $r = 1.05$ ; 2 years is $2 \times 12 = 24$ months, so $n = 24$ . [M1: correct $r$ and $n$ ]\n\nApply the sum formula.\n $S_{24} = \dfrac{u_1(r^{n} - 1)}{r - 1} = \dfrac{20\,(1.05^{24} - 1)}{1.05 - 1}$ [M1: geometric sum formula]\n $S_{24} = \dfrac{20\,(3.2251\ldots - 1)}{0.05} = \dfrac{20 \times 2.2251\ldots}{0.05} = 890.03\ldots$ [A1]\n\nRound for money.\n $S_{24} \approx 890.04$ dollars. [A1]\n\nThis is a SUM, so the series formula is right; using the nth-term formula here would give only the size of the last deposit, not the total.

✓ FULLY WORKED

EXAMPLE 4

Evaluate $\displaystyle \sum_{k=1}^{10} 3(2)^{\,k-1}$ . [3]

1
Recognise the series. Writing out the first few terms, $k = 1$ gives $3(2)^0 = 3$ , $k = 2$ gives $3(2)^1 = 6$ , $k = 3$ gives $3(2)^2 = 12$ . The terms $3, 6, 12, \ldots$ have a constant ratio, so this is a geometric series with $u_1 = 3$ , $r = 2$ and $n = 10$ . [M1: identify $u_1$ , $r$ , $n$ ]\n\nApply the geometric sum formula.\n $S_{10} = \dfrac{u_1(r^{n} - 1)}{r - 1} = \dfrac{3(2^{10} - 1)}{2 - 1}$ [M1: sum formula]\n $S_{10} = \dfrac{3(1024 - 1)}{1} = 3 \times 1023 = 3069.$ [A1]\n\nExpanding two or three terms first is the safe way to tell an arithmetic sigma sum from a geometric one before choosing a formula.

✓ FULLY WORKED

EXAMPLE 5

The first term of a geometric sequence is 5 and the common ratio is 1.2. \n Find the 8th term and the sum of the first 8 terms. [5]

1
Model answer — full working.\n\nThe sequence is geometric with $u_1 = 5$ , $r = 1.2$ .\n\n8th term. Use the nth-term formula with $n = 8$ :\n $u_8 = u_1 r^{\,n-1} = 5 \times 1.2^{\,7}.$ \n $u_8 = 5 \times 3.5831\ldots = 17.9159\ldots \approx 17.9$ (3 s.f.).\n\nSum of the first 8 terms. Use the geometric sum formula with $n = 8$ :\n $S_8 = \dfrac{u_1(r^{n} - 1)}{r - 1} = \dfrac{5\,(1.2^{8} - 1)}{1.2 - 1}.$ \n $S_8 = \dfrac{5\,(4.2998\ldots - 1)}{0.2} = \dfrac{5 \times 3.2998\ldots}{0.2} = 82.4954\ldots \approx 82.5$ (3 s.f.).\n\nAnswers: $u_8 \approx 17.9$ and $S_8 \approx 82.5$ .\n\n---\nHow our marking engine awards the 5 marks:\n\n- M1 — nth-term formula. Awarded for quoting and using $u_n = u_1 r^{\,n-1}$ with $n = 8$ , i.e. writing $5 \times 1.2^{7}$ . It is the method that is rewarded, so this mark survives an arithmetic slip in the value.\n- A1 — value of $u_8$ . Awarded for $u_8 \approx 17.9$ (accepting $17.9159\ldots$ or any correct rounding such as $17.92$ ). This accuracy mark depends on the M1 above.\n- M1 — sum formula. Awarded for quoting the geometric sum formula $S_n = \dfrac{u_1(r^n - 1)}{r - 1}$ . Recognising that a 'sum of the first 8 terms' needs $S_n$ , not $u_n$ , is exactly what this mark checks.\n- M1 — substitution. Awarded for correctly substituting $u_1 = 5$ , $r = 1.2$ , $n = 8$ into that formula: $\dfrac{5(1.2^{8} - 1)}{1.2 - 1}$ . A candidate who writes this line correctly earns the mark even if the final division slips.\n- A1 — value of $S_8$ . Awarded for $S_8 \approx 82.5$ (accepting the unrounded $82.4954\ldots$ or any correct rounding). FT applies: a candidate whose $u_8$ was slightly off but who substitutes and evaluates the sum correctly on their own figures still earns this final mark.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $S_8$ written via the $\dfrac{u_1(1 - r^n)}{1 - r}$ form, and accepts $17.9$ , $17.92$ or the full $17.9159\ldots$ for $u_8$ . Once a correct final value appears, later restatements do not lose marks (ISW).\n\nBottom line: three of the five marks are method marks that survive an arithmetic slip, and the two accuracy marks are shielded by follow-through — but only if the formula and substitution are on the page. A bare ' $u_8 = 17.9$ , $S_8 = 82.5$ ' from the GDC with no working risks all three method marks if a single value is mistyped.

✓ FULLY WORKED

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAArithmetic and geometric sequences and series

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 12

Sequence vs series

TAP OR SPACE TO FLIP

ANSWER

A sequence is an ordered list of terms ( $3, 7, 11, 15, \ldots$ ). A series is the sum of those terms ( $3+7+11+15$ ). A sequence is a list; a series is a total.

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1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
Common difference $d$ : subtract consecutive terms, $d = u_{n+1} - u_n$ . A constant difference is what makes a sequence arithmetic.
✓
nth term: $u_n = u_1 + (n-1)d$ — one term of the sequence.
✓
Sum: $S_n = \frac{n}{2}(2u_1 + (n-1)d)$ — the running total of a series. Use $\frac{n}{2}(u_1 + u_n)$ if the last term is known.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Get a Paper 2 calculation marked: identify the sequence, find the term and the sum with full working

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

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Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Get a Paper 2 calculation marked: identify the sequence, find the term and the sum with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS1.1 Standard form, approximation and percentage error NEXT →1.3 Financial mathematics: compound interest, annuities and amortization

Arithmetic and geometric sequences and series

BY THE END, YOU CAN…

1Distinguish arithmetic sequences (constant difference $d$ ) from geometric sequences (constant ratio $r$ ), and find $u_1$ , $d$ and $r$ from the terms of a sequence.
2Use $u_n = u_1 + (n-1)d$ and $S_n = \frac{n}{2}(2u_1+(n-1)d)$ for arithmetic sequences and series, and $u_n = u_1 r^{n-1}$ and $S_n = \frac{u_1(r^n-1)}{r-1}$ for geometric sequences and series.
3Read, expand and evaluate a finite series written in sigma notation, and convert between a series and its sigma form.
4Model applied situations — salaries, savings plans, loan balances, growth and depreciation — as arithmetic or geometric, choosing the right formula and interpreting the answer in context.
5Use a GDC efficiently (nth-term rules, tables and the finance solver) while still presenting the method so that method marks and follow-through protect the final answer.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Two Ways a Pattern Can Grow

THINK OF IT LIKE…

1
Decide the type. Look at consecutive terms. If you SUBTRACT and always get the same number, it is arithmetic with common difference $d$ . If you DIVIDE and always get the same number, it is geometric with common ratio $r$ .
2
Pull out the parameters. Write down the first term $u_1$ , the step $d$ or ratio $r$ , and the term number $n$ the question asks about. Read percentage changes carefully: a 4% rise means $r = 1.04$ , a 4% fall means $r = 0.96$ .
3
Pick the formula. One term uses the $u_n$ formula; a running total uses the $S_n$ formula. All four are in your formula booklet — the skill is choosing, not memorising.
4
Substitute, compute, interpret. Let the GDC do the arithmetic, then round sensibly and check the answer makes sense for the context — money to 2 decimal places, people to a whole number.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

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

Full topic notes

Formal explanation with the rigour you need for the exam.

Arithmetic sequences: adding a constant step

Sum of the first $n$ terms: $S_n = \dfrac{n}{2}\bigl(2u_1 + (n-1)d\bigr)$ , \n or, when the last term $u_n$ is known, $S_n = \dfrac{n}{2}(u_1 + u_n)$ .

Common difference $d$ : subtract consecutive terms, $d = u_{n+1} - u_n$ . A constant difference is what makes a sequence arithmetic.
nth term: $u_n = u_1 + (n-1)d$ — one term of the sequence.
Sum: $S_n = \frac{n}{2}(2u_1 + (n-1)d)$ — the running total of a series. Use $\frac{n}{2}(u_1 + u_n)$ if the last term is known.

Geometric sequences: multiplying by a constant ratio

Sum of a geometric series

Sigma notation

$\displaystyle \sum_{k=1}^{n} u_k = u_1 + u_2 + \cdots + u_n.$ \n The counter $k$ runs from the lower value to the upper value; each value is substituted into $u_k$ and the results are added.

Common mistakes examiners penalise

Mixing up arithmetic and geometric — always test consecutive terms both ways first. Constant DIFFERENCE means arithmetic ( $d$ ); constant RATIO means geometric ( $r$ ). Using the wrong family of formula loses the whole question.
The off-by-one on $n$ — the exponent and bracket are $n-1$ , not $n$ . The 10th term uses $(10-1) = 9$ steps. And 'after $n$ years' from a starting value is often the $(n+1)$ th term, because the starting value is the 1st term.
Finding $r$ by subtracting — the common ratio is $u_{n+1} \div u_n$ , not $u_{n+1} - u_n$ . Dividing consecutive terms is the only correct way; subtracting gives the difference, which is meaningless for a geometric sequence.
Confusing a term with a total — 'find the 8th term' needs the $u_n$ formula; 'find the total of the first 8 terms' needs the $S_n$ formula. Read whether the question wants one value or a running sum.
Wrong ratio from a percentage — a 15% decrease is $r = 0.85$ , not $0.15$ or $-0.15$ ; a 15% increase is $r = 1.15$ . Convert the percentage to the multiplier that takes 100% to the new value.
Over-rounding mid-calculation — carry the GDC's full figures and round only the final answer, sensibly for the context (money to 2 d.p., people to a whole number, otherwise 3 s.f.).
Writing no method with a GDC answer — a GDC is allowed on every AI paper, but a lone number earns no method mark. Show the formula, the substitution, or the finance-solver inputs so the working is on the page.

Model answer — marked the way our engine marks it

Where this leads

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

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The salary rises by a fixed amount each year, so this is arithmetic with $u_1 = 28000$ and $d = 750$ .\n\n**(a) Salary in the 10th year ( $n = 10$ ).\nUse the nth-term formula:\n $u_{10} = u_1 + (10-1)d = 28000 + 9 \times 750$ [M1: nth-term formula with $n-1 = 9$ ]\n $u_{10} = 28000 + 6750 = 34750.$ \nThe salary in the 10th year is 34 750 dollars. [A1]\n\n(b) Total over the first 10 years.**\nThis is a sum, so use the series formula:\n $S_{10} = \dfrac{10}{2}\bigl(2 \times 28000 + (10-1) \times 750\bigr)$ [M1: sum formula]\n $S_{10} = 5\,(56000 + 6750) = 5 \times 62750 = 313750.$ [A1]\nThe total earned over the first 10 years is 313 750 dollars. [A1]\n\nNotice part (b) uses the SUM formula, not the term formula — a total always calls for $S_n$ .

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EXAMPLE 2

A car is bought for 25 000 dollars and depreciates by 18% each year. \n Find its value, to the nearest dollar, after 5 years. [4]

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The value is multiplied by the same factor each year, so this is geometric.\n\nIdentify the parameters. $u_1 = 25000$ . Losing 18% leaves 82%, so $r = 0.82$ . 'After 5 years', with the purchase price as the 1st term, is the 6th term, so $n = 6$ . [M1: correct $r$ and $n$ ]\n\nApply the nth-term formula.\n $u_6 = u_1 r^{\,6-1} = 25000 \times 0.82^{5}$ [M1: geometric nth-term formula]\n $u_6 = 25000 \times 0.3707\ldots = 9265.58\ldots$ [A1]\n\nRound for the context. Money to the nearest dollar:\n $u_6 \approx 9266$ dollars. [A1]\n\nThe ' $0.82$ ' (not $-0.18$ ) captures the decay, and the ' $n = 6$ ' reflects that the initial value is the first term.

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EXAMPLE 3

A saver deposits 20 dollars in the first month and increases each monthly deposit by 5%. \n Find the total amount deposited after 2 years, to the nearest cent. [4]

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Each deposit is 5% larger than the last, so the deposits form a geometric sequence and their total is a geometric series.\n\nIdentify the parameters. $u_1 = 20$ ; a 5% increase gives $r = 1.05$ ; 2 years is $2 \times 12 = 24$ months, so $n = 24$ . [M1: correct $r$ and $n$ ]\n\nApply the sum formula.\n $S_{24} = \dfrac{u_1(r^{n} - 1)}{r - 1} = \dfrac{20\,(1.05^{24} - 1)}{1.05 - 1}$ [M1: geometric sum formula]\n $S_{24} = \dfrac{20\,(3.2251\ldots - 1)}{0.05} = \dfrac{20 \times 2.2251\ldots}{0.05} = 890.03\ldots$ [A1]\n\nRound for money.\n $S_{24} \approx 890.04$ dollars. [A1]\n\nThis is a SUM, so the series formula is right; using the nth-term formula here would give only the size of the last deposit, not the total.

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EXAMPLE 4

Evaluate $\displaystyle \sum_{k=1}^{10} 3(2)^{\,k-1}$ . [3]

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Recognise the series. Writing out the first few terms, $k = 1$ gives $3(2)^0 = 3$ , $k = 2$ gives $3(2)^1 = 6$ , $k = 3$ gives $3(2)^2 = 12$ . The terms $3, 6, 12, \ldots$ have a constant ratio, so this is a geometric series with $u_1 = 3$ , $r = 2$ and $n = 10$ . [M1: identify $u_1$ , $r$ , $n$ ]\n\nApply the geometric sum formula.\n $S_{10} = \dfrac{u_1(r^{n} - 1)}{r - 1} = \dfrac{3(2^{10} - 1)}{2 - 1}$ [M1: sum formula]\n $S_{10} = \dfrac{3(1024 - 1)}{1} = 3 \times 1023 = 3069.$ [A1]\n\nExpanding two or three terms first is the safe way to tell an arithmetic sigma sum from a geometric one before choosing a formula.

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EXAMPLE 5

The first term of a geometric sequence is 5 and the common ratio is 1.2. \n Find the 8th term and the sum of the first 8 terms. [5]

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Model answer — full working.\n\nThe sequence is geometric with $u_1 = 5$ , $r = 1.2$ .\n\n8th term. Use the nth-term formula with $n = 8$ :\n $u_8 = u_1 r^{\,n-1} = 5 \times 1.2^{\,7}.$ \n $u_8 = 5 \times 3.5831\ldots = 17.9159\ldots \approx 17.9$ (3 s.f.).\n\nSum of the first 8 terms. Use the geometric sum formula with $n = 8$ :\n $S_8 = \dfrac{u_1(r^{n} - 1)}{r - 1} = \dfrac{5\,(1.2^{8} - 1)}{1.2 - 1}.$ \n $S_8 = \dfrac{5\,(4.2998\ldots - 1)}{0.2} = \dfrac{5 \times 3.2998\ldots}{0.2} = 82.4954\ldots \approx 82.5$ (3 s.f.).\n\nAnswers: $u_8 \approx 17.9$ and $S_8 \approx 82.5$ .\n\n---\nHow our marking engine awards the 5 marks:\n\n- M1 — nth-term formula. Awarded for quoting and using $u_n = u_1 r^{\,n-1}$ with $n = 8$ , i.e. writing $5 \times 1.2^{7}$ . It is the method that is rewarded, so this mark survives an arithmetic slip in the value.\n- A1 — value of $u_8$ . Awarded for $u_8 \approx 17.9$ (accepting $17.9159\ldots$ or any correct rounding such as $17.92$ ). This accuracy mark depends on the M1 above.\n- M1 — sum formula. Awarded for quoting the geometric sum formula $S_n = \dfrac{u_1(r^n - 1)}{r - 1}$ . Recognising that a 'sum of the first 8 terms' needs $S_n$ , not $u_n$ , is exactly what this mark checks.\n- M1 — substitution. Awarded for correctly substituting $u_1 = 5$ , $r = 1.2$ , $n = 8$ into that formula: $\dfrac{5(1.2^{8} - 1)}{1.2 - 1}$ . A candidate who writes this line correctly earns the mark even if the final division slips.\n- A1 — value of $S_8$ . Awarded for $S_8 \approx 82.5$ (accepting the unrounded $82.4954\ldots$ or any correct rounding). FT applies: a candidate whose $u_8$ was slightly off but who substitutes and evaluates the sum correctly on their own figures still earns this final mark.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $S_8$ written via the $\dfrac{u_1(1 - r^n)}{1 - r}$ form, and accepts $17.9$ , $17.92$ or the full $17.9159\ldots$ for $u_8$ . Once a correct final value appears, later restatements do not lose marks (ISW).\n\nBottom line: three of the five marks are method marks that survive an arithmetic slip, and the two accuracy marks are shielded by follow-through — but only if the formula and substitution are on the page. A bare ' $u_8 = 17.9$ , $S_8 = 82.5$ ' from the GDC with no working risks all three method marks if a single value is mistyped.

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How it all connects

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MAIN IDEAArithmetic and geometric sequences and series

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QUESTION · 1 / 12

Sequence vs series

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ANSWER

A sequence is an ordered list of terms ( $3, 7, 11, 15, \ldots$ ). A series is the sum of those terms ( $3+7+11+15$ ). A sequence is a list; a series is a total.

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Key takeaways

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Common difference $d$ : subtract consecutive terms, $d = u_{n+1} - u_n$ . A constant difference is what makes a sequence arithmetic.
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nth term: $u_n = u_1 + (n-1)d$ — one term of the sequence.
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Sum: $S_n = \frac{n}{2}(2u_1 + (n-1)d)$ — the running total of a series. Use $\frac{n}{2}(u_1 + u_n)$ if the last term is known.

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Revise Arithmetic and geometric sequences and series for IB Mathematics: Applications and Interpretation (1.2)

In simple terms

Explore the concept

Key formulas

Full topic notes

Arithmetic sequences: adding a constant step

Geometric sequences: multiplying by a constant ratio

Sum of a geometric series

Sigma notation

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 calculation marked: identify the sequence, find the term and the sum with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

Arithmetic sequences: adding a constant step

Geometric sequences: multiplying by a constant ratio

Sum of a geometric series

Sigma notation

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 calculation marked: identify the sequence, find the term and the sum with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Arithmetic and geometric sequences and series