Series Revision Notes — 9709 Paper 1

Sequences, progressions, and telling them apart

Syllabus requirement · §1.6

“

recognise arithmetic and geometric progressions … Including knowledge that numbers a, b, c are ‘in arithmetic progression’ if 2b = a + c (or equivalent) and are ‘in geometric progression’ if b² = ac (or equivalent).

”

A sequence is just a list of numbers in a fixed order:

3,\ 7,\ 11,\ 15,\ 19,\ \ldots

We call the numbers terms. The first term gets the symbol $u_1$ , the second $u_2$ , and in general the $n$ th term is $u_n$ . In this list $u_1 = 3$ , $u_4 = 15$ .

A series is what you get when you add the terms of a sequence: $3 + 7 + 11 + 15 + 19$ . The sum of the first $n$ terms gets the symbol $S_n$ , so here $S_3 = 3 + 7 + 11 = 21$ .

A progression is a sequence built by one repeated rule. The syllabus needs exactly two of them.

The two rules

Look at what happens between consecutive terms.

3,\ 7,\ 11,\ 15,\ 19 \qquad \text{each term is the one before } \mathbf{+\,4}

Nothing is being multiplied; the same number is being added every time. That is an arithmetic progression (AP), and the number you add is the common difference, written $d$ . Here $d = 4$ . Note that $d$ can be negative — $20, 17, 14, 11$ is an AP with $d = -3$ — and it can be a fraction.

3,\ 6,\ 12,\ 24,\ 48 \qquad \text{each term is the one before } \mathbf{\times\,2}

Now nothing is being added; the same number is being multiplied in every time. That is a geometric progression (GP), and the multiplier is the common ratio, written $r$ . Here $r = 2$ . Again $r$ can be negative ( $3, -6, 12, -24$ has $r = -2$ ) and it is very often a fraction.

The letter $a$ always means the first term of whichever progression you are in.

An arithmetic progression grows by the same amount at every step, so the bars rise evenly. A geometric one is multiplied by the same factor at every step, so each rise is bigger than the one before.

How to test a list you are handed

Take the first three terms $u_1, u_2, u_3$ and do both checks:

Differences. Is $u_2 - u_1$ equal to $u_3 - u_2$ ? If yes, it is arithmetic and that number is $d$ .
Ratios. Is $\dfrac{u_2}{u_1}$ equal to $\dfrac{u_3}{u_2}$ ? If yes, it is geometric and that number is $r$ .

Try it on $1,\ 4,\ 9,\ 16$ : the differences are $3, 5, 7$ — not equal, so not arithmetic. The ratios are $4$ and $2.25$ — not equal, so not geometric. It is neither; it is the square numbers, and none of the formulae on this page apply to it.

The three-term conditions, and where they come from

Exam questions rarely hand you numbers. They hand you expressions — "the first three terms of a geometric progression are $25$ , $4q-1$ and $13-q$ " — and expect you to turn the words "are in geometric progression" into an equation.

You could do that with the ratio test, but there is a tidier form. Suppose $a$ , $b$ , $c$ are three consecutive terms.

If they are arithmetic, the two gaps are equal:

b - a = c - b

Add $b$ to both sides and add $a$ to both sides:

2b = a + c

If they are geometric, the two ratios are equal:

\frac{b}{a} = \frac{c}{b}

Cross-multiply:

b^{2} = ac

That is all they are — the difference test and the ratio test, rearranged so there are no fractions to trip over.

Three consecutive terms a, b, c

2b = a + c

…are in arithmetic progression

b^{2} = ac

…are in geometric progression

d = b - a = c - b

Common difference, either gap

r = \dfrac{b}{a} = \dfrac{c}{b}

Common ratio, either ratio

“Middle term squared equals the outer two multiplied”

$b^{2} = ac$ is worth saying out loud in words, because in the exam the three terms will be ugly expressions and you need to know which one is the middle. For $25$ , $4q-1$ , $13-q$ it gives

(4q-1)^{2} = 25(13-q)

directly, with no fractions anywhere. That single line is usually the first mark.

Demonstration on invented numbers

Before touching a past paper, watch the machinery work on numbers chosen to be clean.

Suppose the first three terms of an arithmetic progression are $x$ , $2x+1$ and $x^{2}-2$ . Find $x$ .

The words "arithmetic progression" mean $2b = a + c$ , with $a = x$ , $b = 2x+1$ , $c = x^{2}-2$ :

2(2x+1) = x + \left(x^{2}-2\right)

Expand the left-hand side:

4x + 2 = x + x^{2} - 2

Collect everything on the side that keeps $x^{2}$ positive:

0 = x^{2} + x - 2 - 4x - 2

0 = x^{2} - 3x - 4

Factorise:

(x-4)(x+1) = 0 \quad\Longrightarrow\quad x = 4 \ \text{ or } \ x = -1

Now check both roots against the question, which is the step students skip.

$x = 4$ gives the terms $4,\ 9,\ 14$ . Differences $5$ and $5$ . A genuine AP with $d = 5$ . ✓
$x = -1$ gives the terms $-1,\ -1,\ -1$ . Differences $0$ and $0$ . Technically an AP, but a constant one with $d = 0$ .

If the question had said "the common difference is not zero" — and questions do say exactly that — the second root would have to be rejected. That is the whole point of the sentence.

Now the same thing for a geometric progression, so you can see the quadratic arrive from the other condition.

The first three terms of a geometric progression are $x$ , $6$ and $x+5$ . Find the possible values of $x$ .

Here $b^{2} = ac$ gives

6^{2} = x(x+5)

36 = x^{2} + 5x

x^{2} + 5x - 36 = 0

(x+9)(x-4) = 0 \quad\Longrightarrow\quad x = 4 \ \text{ or } \ x = -9

Check both:

$x = 4$ : terms $4,\ 6,\ 9$ , so $r = \tfrac{6}{4} = \tfrac32$ , and $6 \times \tfrac32 = 9$ . ✓
$x = -9$ : terms $-9,\ 6,\ -4$ , so $r = \tfrac{6}{-9} = -\tfrac23$ , and $6 \times -\tfrac23 = -4$ . ✓

Both are real geometric progressions. A negative common ratio is perfectly legal — it just makes the terms alternate in sign. Unless the question tells you $x$ is positive, or that the progression converges, or that $r$ is negative, you must give both.

Turning “are in … progression” into an answer

1
Write down which condition applies. Arithmetic $\Rightarrow 2b = a+c$ . Geometric $\Rightarrow b^{2} = ac$ . Identify $a$ , $b$ , $c$ explicitly — the middle expression is $b$ .
This one line is almost always worth a mark on its own, even if everything after it goes wrong.
2
Expand both sides and collect into a three-term quadratic $= 0$ . Do not cancel anything yet.
Mark schemes award a mark for “reaching a 3-term quadratic”. Cancelling an x too early can lose a root.
3
Solve it — factorise if you can, otherwise the formula. Show the solving method: several mark schemes say "SC B1 if no method shown for solving the quadratic", meaning a bare answer costs you.
4
Read the question again and filter the roots. Look for the words positive, negative, not zero, convergent, all terms are positive. Each of those exists solely to kill one root.
Examiners write these conditions in because they know two roots come out. Ignoring them is the standard way to lose the final A mark.
5
Feed the surviving root back in and answer what was actually asked — usually a term or a sum, not $x$ itself.

Three expressions in geometric progression

9709/13 O/N 2023 Q5(a)3 marks

The first, second and third terms of a geometric progression are $2p + 6$ , $5p$ and $8p + 2$ respectively.

Find the possible values of the constant $p$ .

Show full working

1
"Geometric" means $b^{2} = ac$ . The middle term is $b = 5p$ , and the outer two are $a = 2p + 6$ and $c = 8p + 2$ : $(5p)^{2} = (2p+6)(8p+2)$
Name the middle term first. Squaring the wrong expression is the usual way this line goes wrong.
2
Left-hand side: $(5p)^{2} = 25p^{2}$ — the 5 is squared as well as the $p$ .
3
Right-hand side, one product at a time: $(2p+6)(8p+2) = 16p^{2} + 4p + 48p + 12 = 16p^{2} + 52p + 12$
4
Set them equal: $25p^{2} = 16p^{2} + 52p + 12$
5
Collect everything on the left, keeping the $p^{2}$ term positive: $9p^{2} - 52p - 12 = 0$
This three-term quadratic is itself a method mark, so write it down before solving.
6
Factorise. Two numbers multiplying to $9 \times (-12) = -108$ and adding to $-52$ are $-54$ and $2$ : $9p^{2} - 54p + 2p - 12 = 9p(p-6) + 2(p-6) = (9p+2)(p-6) = 0$
7
So $p = 6 \qquad\text{or}\qquad p = -\frac29$
8
Check both. $p = 6$ gives $18,\ 30,\ 50$ : ratio $\tfrac{30}{18} = \tfrac53$ and $30 \times \tfrac53 = 50$ ✓. $p = -\tfrac29$ gives $\tfrac{50}{9},\ -\tfrac{10}{9},\ \tfrac29$ : ratio $-\tfrac15$ and $-\tfrac{10}{9} \times \left(-\tfrac15\right) = \tfrac29$ ✓.
Nothing in the question restricts p, so both values are answers. Only reject a root when the question gives you a reason.

Answer

$p = 6$ or $p = -\tfrac29$

Compare this with the invented demonstrations above: the same three moves — write $b^{2} = ac$ , reach a three-term quadratic, solve — and then read the question for a condition. Here there is none, so both roots stand.

Common mistakes

For terms $a,\ b,\ c$ in GP, writing $b^{2} = a + c$
$b^{2} = ac$ for geometric; $2b = a + c$ for arithmetic
The two conditions get swapped constantly under pressure. Geometric multiplies, so its condition multiplies.
$a^{2} - 3a = 0 \Rightarrow a = 3$
$a(a-3) = 0 \Rightarrow a = 0$ or $a = 3$
Dividing an equation by the unknown deletes a root. Factorise instead — then reject on the question's own words.
Giving both roots when the question says “ $q$ is a positive constant”
Rejecting the root that breaks the stated condition, and saying so
Every ‘positive’, ‘negative’, ‘non-zero’ and ‘convergent’ in a stem is there to remove exactly one root.
Finding $x$ and stopping
Substituting $x$ back and answering what was asked — the 10th term, or the sum of the first 50 terms
Finding the unknown is usually only the middle of the question. The last marks are for using it.
Treating $1,\ 4,\ 9,\ 16$ as a progression and reaching for $u_n = a+(n-1)d$
Checking the differences and the ratios first — this one is neither
Only APs and GPs are on the syllabus. If a list is neither, the question is asking something else.

Your turn — recognising

Work each one to a final number. The habit to build is the last step: check your root against the words in the question before writing the answer down.

1
Decide whether each list is arithmetic, geometric or neither. Where it is a progression, state $a$ and $d$ or $a$ and $r$ .

(i) $5,\ 9,\ 13,\ 17$ (ii) $5,\ 10,\ 20,\ 40$ (iii) $2,\ 6,\ 12,\ 20$ (iv) $81,\ -27,\ 9,\ -3$
Stuck? Show hint
For each list write the three differences underneath, then the three ratios. Only one of those rows will be constant — if neither is, it is neither.
Show solution
1. 1
  (i) Differences: $9-5 = 4$ , $13-9 = 4$ , $17-13 = 4$ . Constant. Arithmetic, $a = 5$ , $d = 4$ .
2. 2
  (ii) Differences: $5, 10, 20$ — not constant. Ratios: $\tfrac{10}{5} = 2$ , $\tfrac{20}{10} = 2$ , $\tfrac{40}{20} = 2$ . Constant. Geometric, $a = 5$ , $r = 2$ .
3. 3
  (iii) Differences: $4, 6, 8$ — not constant. Ratios: $3, 2, \tfrac{20}{12} = \tfrac53$ — not constant. Neither.
  These are the numbers n(n+1). A perfectly good sequence, but no Paper 1 formula applies to it.
4. 4
  (iv) Ratios: $\tfrac{-27}{81} = -\tfrac13$ , $\tfrac{9}{-27} = -\tfrac13$ , $\tfrac{-3}{9} = -\tfrac13$ . Constant. Geometric, $a = 81$ , $r = -\tfrac13$ .
  A negative ratio makes the signs alternate. It is still a perfectly ordinary GP.
Answer
(i) AP, $a=5$ , $d=4$ · (ii) GP, $a=5$ , $r=2$ · (iii) neither · (iv) GP, $a=81$ , $r=-\tfrac13$
29709/12 M/J 2022 Q4(a)2 marks
The first, second and third terms of an arithmetic progression are $k$ , $6k$ and $k + 6$ respectively.

Find the value of the constant $k$ .
Stuck? Show hint
Use $2b = a + c$ with the middle term $6k$ . This time the equation is linear, not quadratic.
Show solution
1. 1
  Apply $2b = a + c$ with $a = k$ , $b = 6k$ , $c = k + 6$ : $2(6k) = k + (k + 6)$
2. 2
  Simplify each side: $12k = 2k + 6$
3. 3
  Subtract $2k$ from both sides: $10k = 6$
4. 4
  Divide by $10$ : $k = \frac{6}{10} = 0.6$
5. 5
  Check: the terms are $0.6,\ 3.6,\ 6.6$ , with gaps $3$ and $3$ ✓.
  Writing the three numbers out takes seconds and confirms the answer.
Answer
$k = 0.6$
39709/11 M/J 2023 Q6(a)3 marks
The first three terms of an arithmetic progression are $\dfrac{p^{2}}{6}$ , $2p-6$ and $p$ .

Given that the common difference of the progression is not zero, find the value of $p$ .
Stuck? Show hint
Use $2b = a+c$ , then multiply everything by 6 to clear the fraction before you try to factorise.
Show solution
1. 1
  Apply $2b = a + c$ with $a = \dfrac{p^{2}}{6}$ , $b = 2p-6$ , $c = p$ : $2(2p-6) = \frac{p^{2}}{6} + p$
2. 2
  Expand the left-hand side: $4p - 12 = \frac{p^{2}}{6} + p$
3. 3
  Multiply every term by $6$ to clear the fraction: $24p - 72 = p^{2} + 6p$
  Multiply every term, including the ones that had no fraction. Missing one is where this question goes wrong.
4. 4
  Collect on the side that keeps $p^{2}$ positive: $0 = p^{2} + 6p - 24p + 72 = p^{2} - 18p + 72$
5. 5
  Factorise — look for two numbers multiplying to $72$ and adding to $-18$ , namely $-6$ and $-12$ : $(p-6)(p-12) = 0 \quad\Longrightarrow\quad p = 6 \ \text{ or } \ p = 12$
6. 6
  Now use the condition. Test $p = 6$ : the terms are $\dfrac{36}{6} = 6$ , $\ 2(6)-6 = 6$ , $\ 6$ . That is $6,\ 6,\ 6$ , so $d = 0$ — rejected.
  This is exactly what “the common difference is not zero” was put there for. Test each root against it explicitly.
7. 7
  Test $p = 12$ : the terms are $\dfrac{144}{6} = 24$ , $\ 2(12)-6 = 18$ , $\ 12$ . Gaps of $-6$ and $-6$ , so $d = -6 \neq 0$ . ✓
Answer
$p = 12$

In the exam

37 parts · 130 marks · 3.5 marks each, 2021–2025

"Recognising" sounds like a one-mark skill, but it is rarely asked on its own. The usual design is: the condition $2b = a + c$ or $b^{2} = ac$ produces a quadratic, the question's wording removes one root, and the surviving value is then pushed through a sum or a sum to infinity. Expect to use all three steps in one question.

Practise recognising progressionsReal past-paper questions · Recognising arithmetic and geometric progressions

Checking your access…

Can you do all of these?

Decide from raw terms whether a list is arithmetic, geometric or neither
Turn " $a$ , $b$ , $c$ are in geometric progression" into $b^2 = ac$ and solve the quadratic that follows
Reject a root because the question said "positive", "negative", "non-zero" or "convergent"
Find $a$ and $d$ from two facts, then answer what was actually asked
Find the smallest $n$ for which a term is negative, rounding the inequality the right way
Sum the terms of an AP that lie between two given values, counting inclusively
Find the least $n$ for which $S_n$ passes a value, by solving a quadratic inequality in $n$
Find $r$ from two non-adjacent terms by dividing and taking the appropriate root
Use $S_n = \dfrac{a(1-r^n)}{1-r}$ , and switch to the $\dfrac{a(r^n-1)}{r-1}$ form when $r > 1$
State $|r| < 1$ unprompted, and use it to reject a root of a quadratic in $r$
Compute $S_\infty = \dfrac{a}{1-r}$ , and set it equal to a given value to find $a$ or $r$
Find the sum to infinity of the terms left after removing some, or of every third term (ratio $r^3$ )
Expand $(a+b)^n$ using Pascal's triangle for small $n$ and $^{n}C_{r}$ for large $n$
Write the general term $\binom{n}{r}a^{n-r}b^{r}$ and pick out one coefficient without expanding
Find the term independent of $x$ by setting the total power of $x$ to zero
Find the coefficient of $x^k$ in a product of two brackets by collecting every pairing, including pairings with negative powers
Set a coefficient equal to a given number and solve the resulting equation for a constant
Divide consecutive trigonometric terms, cancel, and get a simple $r$ such as $2\sin\theta$ ; use exact trigonometric values when the angle is given
Substitute $s = \sin\theta$ , solve the quadratic, and convert back with $\sin^{-1}$
Write the 2nd, 5th and 11th terms of an AP and impose $b^2 = ac$ on them
Factorise rather than divide by $d$ , then reject $d = 0$ using the stated condition

Now do the questions

146 real Paper 1 parts from 2021–2025, sorted by difficulty, with mark schemes

Trigonometry

Differentiation

Series

Sequences, progressions, and telling them apart

The two rules

The three-term conditions, and where they come from

Demonstration on invented numbers

Arithmetic progressions

Geometric progressions

Can you do all of these?

Sequences, progressions, and telling them apart

The two rules

The three-term conditions, and where they come from

Demonstration on invented numbers

The rest of this note

Can you do all of these?