Sequences, Recurrence Relations and Arithmetic Series

~50 min · WMA12 · 4.1

WMA12 · 4.1 · 50 min

A sequence can be handed to you two structurally different ways — a formula that computes any term directly, or a rule, $x_{n+1} = f(x_n)$ , that only tells you how to get the next term from the one you already have — and almost every mark lost on this topic comes from treating the second kind like the first: applying a rule one time too many or too few, because "the 4th term" and "4 applications of the rule" are not the same count. Arithmetic sequences are the one case where both descriptions collapse into the same clean formula, $u_n = a + (n-1)d$ , and the proof of where that formula — and its partner for the sum, $S_n = \frac{1}{2}n[2a+(n-1)d]$ — actually comes from is not optional background reading. The specification says outright that you are expected to be able to produce it.

Key terms in this lesson

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

Two ways to describe a sequence — a formula, and a rule for the next term

Spec 4.1 names both. The first is a formula for the nth term: a rule like $u_n = 3n + 1$ that computes any term directly from its position, with no need to know any of the others — plug in $n = 50$ and $u_{50}$ is immediate. The second is a , written by the specification itself as $x_{n+1} = f(x_n)$ : a rule for producing the next term from the current one. A recurrence on its own generates nothing — it needs a starting value, $x_1$ , to have something to apply itself to. $x_{n+1} = x_n + 5$ is not a sequence; $x_1 = 2$ , $x_{n+1} = x_n + 5$ is.

Notice the letter switch: the specification writes the general recurrence with $x$ , and writes arithmetic-specific formulae (spec 4.2, and the formula booklet) with $u$ . That is not this course normalising two sources into one convention — it is what the spec itself does, because 4.1 is deliberately general (the function $f$ could be anything: $x_{n+1} = x_n^2$ , $x_{n+1} = \frac{1}{x_n}$ , $x_{n+1} = 3x_n - 1$ ), while 4.2 narrows to one specific, well-behaved case with its own dedicated letter and dedicated formula. Both notations mean the same thing structurally — a rule plus a starting value — and exam questions use whichever letter the question itself introduces.

Generating terms from a recurrence is direct substitution, repeated: given $x_1$ , compute $x_2 = f(x_1)$ , then $x_3 = f(x_2)$ , and so on. The one place this goes wrong is counting — reaching the nth term from the first takes exactly $(n-1)$ applications of $f$ , not $n$ , because $x_1$ is already given for free and needs no application of the rule to exist. Miscounting this by one is the single most common error type in this whole topic, on both sides of the count (applying the rule once too often or stopping one term short), and it is worth writing every intermediate term out explicitly — $x_1, x_2, x_3, \ldots$ — rather than trying to track the count mentally.

Arithmetic sequences, and three families of behaviour

An is the specific case where the recurrence has one particular shape: $u_{n+1} = u_n + d$ for some fixed constant $d$ , the . Every term is obtained from the last by adding the same fixed amount — not multiplying, not squaring, not anything else, just adding $d$ , every single time. That single constraint is what makes the whole topic solvable with clean formulae rather than needing the general machinery of 4.1.

Spec 4.3 names three ways a sequence can behave, independent of whether it happens to be arithmetic. A sequence is increasing if $u_{n+1} > u_n$ for every $n$ — every term is strictly bigger than the one before, with no exceptions anywhere in the sequence. It is decreasing if $u_{n+1} < u_n$ for every $n$ , the mirror condition. And it is periodic with period $k$ if $u_{n+k} = u_n$ for every $n$ — the sequence repeats its exact pattern of values every $k$ terms, forever, not just once by coincidence partway through. An arithmetic sequence with $d > 0$ is always increasing and one with $d < 0$ is always decreasing (the constant addition never lets the direction reverse); an arithmetic sequence with $d = 0$ is constant, which is periodic with period 1 but not increasing or decreasing under the strict inequalities above. A periodic sequence, by contrast, is almost never arithmetic — a genuinely repeating pattern needs the terms to come back down as often as they go up, which a fixed constant addition cannot do unless that addition is exactly zero.

These three labels are properties of the whole sequence, checked for every $n$ — not a description of what happens to look true for the first few terms you happen to write down. A sequence that increases for its first five terms and then decreases is neither increasing nor decreasing by this definition; it is simply a sequence that is not monotonic. Confirming "increasing" or "decreasing" rigorously means showing the defining inequality holds in general, algebraically, not showing it holds for a handful of examples — the same discipline a proof by exhaustion (spec 1.2 of this unit) explicitly forbids in the other direction: checking finitely many cases proves nothing about a claim that has to hold for every $n$ .

Mechanism

Why $u_n = a + (n-1)d$ — it falls out of the recurrence, it is not a separate rule to learn

Start from the recurrence itself, $u_{n+1} = u_n + d$ , with first term $u_1 = a$ , and apply it repeatedly rather than accepting the closed formula as given. $u_2 = u_1 + d = a + d$ . $u_3 = u_2 + d = (a+d) + d = a + 2d$ . $u_4 = u_3 + d = (a+2d) + d = a + 3d$ . The pattern is exactly what "add $d$ every step" should produce: to reach $u_n$ from $u_1$ takes $(n-1)$ steps, each adding one more $d$ , so $u_n = a + (n-1)d$ . This is not a coincidence discovered by trying a few cases and guessing the pattern — it is a direct consequence of what "arithmetic" was defined to mean in the first place, and it is the same $(n-1)$ -not- $n$ counting from the recurrence-relation teach block above, now packaged into a formula so that it never has to be counted by hand again. Two consequences follow directly. First, this is exactly why $u_1$ corresponds to $(n-1) = 0$ rather than $(n-1) = 1$ : the first term needs zero applications of the recurrence to exist, and the formula has to agree with that or it would be wrong at its own starting point. Second, an arithmetic sequence is fully determined by exactly two numbers, $a$ and $d$ — every other term is a deterministic consequence of those two, which is the reason a question that gives you two arbitrary terms of an arithmetic sequence (not necessarily the first two) is always secretly a simultaneous-equations question: two unknowns, two given data points, one linear system.

Mechanism

Why the sum formula works — a symmetry, not a fact to memorise

Spec 4.2 states plainly that the proof of $S_n = \frac{1}{2}n[2a+(n-1)d]$ should be known, not merely the formula itself — and the January 2020 examiner report on exactly this proof found it "surprisingly unusual to see a well set out, fully correct proof", with candidates struggling from the very first line. The proof itself is one idea, carried out once. Write the sum out in full, forwards: $S_n = a + (a+d) + (a+2d) + \cdots + (a+(n-1)d)$ — note the last term is $a + (n-1)d$ , not $a + nd$ , for exactly the reason the mechanism above just established: reaching the nth term takes $(n-1)$ steps from the first. Now write the identical sum a second time, in reverse order: $S_n = (a+(n-1)d) + (a+(n-2)d) + \cdots + (a+d) + a$ . Nothing has changed — it is the same set of numbers, added in the opposite order, so it is still equal to $S_n$ . The move that makes the proof work is adding these two versions together, term by term, in their current positions: the first term of the first version pairs with the first term of the second version, and so on down the line. The first pairing is $a$ with $a + (n-1)d$ , summing to $2a + (n-1)d$ . The second pairing is $(a+d)$ with $(a+(n-2)d)$ , summing to $2a + (n-1)d$ again — the $+d$ on one side is exactly cancelled by the $-d$ on the other, because as you step forward through the first version you step backward through the second, by the same amount, every time. Every pairing produces the same total, $2a+(n-1)d$ , and there are $n$ of them — one for each of the $n$ terms in the original sum. So adding the two versions gives $2S_n = n[2a+(n-1)d]$ , and dividing by 2 gives the formula. Nobody had to guess $2a+(n-1)d$ ; it is what falls out of pairing the smallest term with the largest, the second-smallest with the second-largest, and so on, and finding — because the sequence moves by a constant step — that every pair happens to land on the same total. This is also, structurally, the identical trick behind the more famous version of it: summing the first $n$ natural numbers, $1 + 2 + \cdots + n$ , is exactly this proof with $a=1$ and $d=1$ , giving $S_n = \frac{1}{2}n(n+1)$ — the specific case the spec names separately ("the sum of the first n natural numbers") precisely because it is common enough to deserve its own recognition, not because it needs a different argument.

Worked, in full

Two given terms, secretly a simultaneous-equations question — finding $a$ , $d$ , and then $S_{20}$

01
An arithmetic sequence has $u_5 = 33$ and $u_{12} = 68$ . Translate both into equations using $u_n = a + (n-1)d$ : $u_5 = a + 4d = 33$ and $u_{12} = a + 11d = 68$ . Two given terms of an arithmetic sequence are always two linear equations in the two unknowns $a$ and $d$ — the structure the question hides is exactly this, whatever cover story surrounds it.
Earns: M1 — forms two correct linear equations in $a$ and $d$ from the two given terms, using the correct $(n-1)$ power of $d$ for each (4 for the 5th term, 11 for the 12th).
02
Subtract the first equation from the second to eliminate $a$ , since $a$ appears identically in both: $(a+11d) - (a+4d) = 68 - 33$ , so $7d = 35$ , giving $d = 5$ .
Earns: M1 — a correct method for eliminating one unknown from the pair of equations, dependent on both equations from stage 1 being correctly formed.
03
Substitute $d = 5$ back into either original equation: $a + 4(5) = 33$ , so $a = 13$ . Check against the other equation, which was not used to find this value: $13 + 11(5) = 13 + 55 = 68$ . ✓
Earns: A1 — both $a$ and $d$ correct. The check against the unused equation is not itself a mark-scheme line, but it costs one line of arithmetic and would have caught a slip immediately, before it propagated into every later part of the question.
04
Substitute $a = 13$ , $d = 5$ , $n = 20$ into the sum formula: $S_{20} = \frac{1}{2}(20)[2(13) + 19(5)] = 10[26 + 95] = 10(121) = 1210$ .
Earns: M1 A1 — method mark for correct substitution of the newly-found $a$ and $d$ into the sum formula, earned on their own stage-3 values with no 'ft' qualifier needed (M marks are for 'knowing a method and attempting to apply it', which does not depend on whether the values applied are themselves correct — 'ft' attaches only to A and B marks, never to M marks — so an arithmetic slip in stage 3 would still leave this method mark available), and the accuracy mark for the correct final value.

Complete it yourself

Complete the chain — find $a$ , $d$ and $S_{15}$ given $u_3 = 20$ and $u_8 = 50$

01
Translate the two given terms into equations using $u_n = a + (n-1)d$ : $u_3 = a + 2d = 20$ and $u_8 = a + 7d = 50$ .
02
Subtract to eliminate $a$ : $(a+7d) - (a+2d) = 50 - 20$ , so $5d = 30$ , giving $d = 6$ .

Same question, every valid method

An arithmetic series has first term 3 and common difference 2. Given that the sum of the first $n$ terms is 195, find the value of $n$ . (VERIDIAN-original question — not a reproduction of any past-paper question.)

2 valid methods · every one reaches $n = 13$ · 3 marks available

01
$S_n = \frac{1}{2}n[2(3)+(n-1)(2)] = \frac{1}{2}n[2n+4] = n^2+2n$ . Setting $S_n = 195$ : $n^2 + 2n - 195 = 0$ .
Method mark for correctly forming a three-term quadratic equation in $n$ by substituting $a$ and $d$ into the sum formula and setting the result equal to the given total. A sum-equals-a-target-value question is, underneath, exactly the same quadratic-solving skill covered in the quadratics topic — the unknown just happens to be called $n$ instead of $x$ .
M1
02
$(n-13)(n+15) = 0$ , so $n = 13$ or $n = -15$ .
Accuracy mark for both roots of the quadratic, found by factorisation. The general principles' condition for the method mark on a three-term quadratic applies unchanged here: the brackets must multiply back to the correct leading coefficient (1) and correct constant ( $-195$ ), which $(n-13)(n+15)$ does, since $(-13)(15) = -195$ .
A1
03
$n$ counts a number of terms, so it must be a positive integer: $n = -15$ is rejected, leaving $n = 13$ .
Final accuracy mark, and the one this question type actually tests. A quadratic in $n$ always produces two mathematical roots, but only one of them can answer a question about how many terms a series has — dropping this line is the arithmetic-series version of forgetting a domain restriction on a parameter question.
A1

Fastest when a factor pair is visible by inspection: $195 = 13 \times 15$ , with the two factors 2 apart, is exactly the kind of clean split worth trying first once the quadratic is written down. As always, the risk is searching for factors that are not there — if nothing clean turns up within a few tries, switch to the formula rather than keep guessing.

Marked, line by line

(a) An arithmetic series has first term $a$ and common difference $d$ . Prove that the sum of the first $n$ terms is given by $S_n = \frac{1}{2}n[2a+(n-1)d]$ . (3) (b) A different arithmetic series has first term 8 and common difference $-3$ . Find $S_{20}$ , the sum of the first 20 terms of this series. (3) — Part (a) is this exact question, verbatim: January 2020 Q8(i) (WMA12_01, `P60571RA`), 3 marks as printed on the paper — B1 M1 A1*, confirmed against the real mark scheme (`WMA12_01_2001_MS`) via both `pdftotext -layout` and `-raw` extraction, which agree exactly. Part (b) is an original numerical follow-up, not a reproduction of any past-paper question — no real mark scheme exists for it, and its marks are modelled on verified general conventions per the flag block below.

6 marks available

(a) — 3 marks

01
$S_n = a + (a+d) + (a+2d) + \cdots + (a+(n-2)d) + (a+(n-1)d)$
Independent mark for an expression for Sₙ in a minimum of 3 terms in a and d, including the first and last terms, with the correct final term a + (n − 1)d — not a + nd. This is precisely the line the January 2020 examiner report (Q8(i), the real question this part reproduces) flags as the most commonly wrong part of this exact proof: 'a surprising proportion of candidates struggled to write a correct expression of Sₙ at the start of their proof, often having the last term incorrect.'
B1
02
Reversing the order and adding term-by-term to the original: $S_n = (a+(n-1)d) + (a+(n-2)d) + \cdots + (a+d) + a$ , added to $S_n = a + (a+d) + \cdots + (a+(n-1)d)$ .
Method mark for attempting to reverse the sum and add it to the original, with at least two matching pairs shown to establish the pattern — the real mark scheme's own wording ('Attempts to reverse their sum and add terms. Must include at least two pairs of matching terms...') covers reversing and adding together as ONE method mark, not two; the same report records this exact step as sometimes skipped entirely: 'Others failed to reverse the terms in the sum correctly, or at all.'
M1
03
Every pair sums to $a + (a+(n-1)d) = 2a+(n-1)d$ , the same value each time, and there are $n$ such pairs, so $2S_n = n[2a+(n-1)d]$ . Dividing by 2: $S_n = \frac{1}{2}n[2a+(n-1)d]$ , as required.
Accuracy mark, answer given (ag/cso) — the real mark scheme's own wording is 'Correctly achieves the given result including the intermediate line 2S=n{2a+(n-1)d}', ONE mark covering both reaching that intermediate line and the final division, with no separate mark for either half. Because the result is printed on the question, this mark requires every line above to be correct and complete, with at least 3 terms shown for the sum and its reverse — a proof that reaches a different-looking expression, even one algebraically equal to the printed form, does not earn it without also showing that equivalence explicitly.
A1*

(b) — 3 marks

101
$S_n = \frac{1}{2}n[2a+(n-1)d]$ , with $a=8$ , $d=-3$ , $n=20$ : $S_{20} = \frac{1}{2}(20)[2(8)+19(-3)]$
Method mark for correct substitution into the sum formula with all three values — a, d and n — correctly identified.
M1
102
$= 10[16-57] = 10(-41)$
Correct simplification of the bracket. 19 × (−3) = −57 — a genuine place to lose the mark by mishandling the sign of a negative common difference.
A1
103
$S_{20} = -410$
cao — correct answer only. A negative total is a legitimate, expected result once d is negative and enough terms have been summed for the negative terms to dominate — it is not, on its own, a sign that a mistake was made.
A1

In your own words

In one sentence: why does proving $S_n = \frac{1}{2}n[2a+(n-1)d]$ require writing the series twice — once forwards, once in reverse — rather than just adding the $n$ terms in the order they were written?

Named traps

sum-proof-last-term-off-by-one: Confirmed directly on the standard sum-formula proof: "a surprising proportion of candidates struggled to write a correct expression of Sₙ at the start of their proof, often having the last term incorrect" (Jan 2020, Q8(i)). The fix is the same mechanism taught earlier in this lesson: reaching the nth term from the first takes (n − 1) steps of the recurrence, so the last term of an n-term series is a + (n − 1)d, never a + nd.
sum-proof-reversal-skipped: Confirmed on the same question: "Others failed to reverse the terms in the sum correctly, or at all" (Jan 2020, Q8(i)). Without the reversed copy, there is nothing for the forward copy to pair against, and the whole argument — every pair summing to the same constant — has no pairs to point at. This step is not optional decoration on the proof; it is the entire mechanism.
sum-proof-insufficient-terms-shown: Confirmed on the same question: candidates "did not display a sufficient number of terms to make subsequent steps in their proof to complete" (Jan 2020, Q8(i)). Two terms at each end plus an ellipsis — $a + (a+d) + \cdots + (a+(n-2)d) + (a+(n-1)d)$ — is the minimum that actually shows the pattern; a single "..." with no visible terms either side of it gives an examiner nothing to check the pairing against.
induction-attempted-for-standard-proof: Confirmed on the same question: "Attempts at proof by induction were also seen in a few cases, though little progress was made in most of these" (Jan 2020, Q8(i)). Induction can, in principle, prove this result, but it is not the proof the specification names, and the report records it as the less successful choice in practice. Where a question asks to "prove the standard formula," the pairing argument is the direct, expected route.
sum-proof-assumes-the-result: Confirmed on the same question: "Some candidates either started with, or used something that they were trying to prove, e.g. starting with $S_n = \frac{1}{2}n(a+L)$ , or attempting to use a summation formula that is a generalisation of the result they needed to show" (Jan 2020, Q8(i)). This scores zero on an "ag" (answer given) proof no matter how cleanly the rest of the working is presented, because it assumes the very fact being proved rather than deriving it — a "prove that" question can never legitimately start from a form of its own target result. The only safe starting point is the definition of the series itself, written out term by term, exactly as the mechanism block above does.
sum-proof-pairing-fails-for-odd-n: Confirmed on the same question: some candidates instead tried pairing terms within a single copy of the sum — first-plus-last, second-plus-second-to-last, and so on — "but they did not give due consideration to what happens when there are an odd number of terms, and were unable to score the accuracy mark as a result" (Jan 2020, Q8(i)). An odd number of terms leaves one unpaired middle term this method has no clean way to handle, which is exactly why the write-forwards/write-backwards/add version taught in this lesson is the safer standard route — it pairs the full sum against a full reversed copy of itself, so it never needs $n$ to be even and never has a leftover term to account for separately.
arithmetic-and-geometric-formulas-confused: Confirmed as a recurring, named error category across at least two series (June 2019 and October 2020): candidates reach for the wrong series formula mid-question — applying the geometric sum formula to an arithmetic series, or the reverse — because the two topics sit close together and use similarly-shaped notation. The one-line check that catches it before it starts: an arithmetic sequence has a constant DIFFERENCE between consecutive terms ( $u_{n+1} - u_n = d$ , the same value every time); a geometric one has a constant RATIO instead. Confirm which quantity is actually constant before reaching for either sum formula.

Retrieval — with feedback on every choice

Question 1

2 marks

An arithmetic sequence has $u_1 = 6$ and $u_4 = 18$ . What is the common difference $d$ ?

Question 2

2 marks

A sequence is defined by $u_1 = 3$ and $u_{n+1} = 2u_n - 1$ . What is the value of $u_3$ ?

Question 3

1 mark

The sequence defined by $u_n = (-1)^n$ is . What is its period?

Question 4

3 marks

An arithmetic series has first term $10$ and common difference $-2$ . Find $S_{25}$ .

Question 5

2 marks

In the standard proof that $S_n = \frac{1}{2}n[2a+(n-1)d]$ , why is the series written a second time in reverse order before the two versions are added?

Reference — not a study method, a lookup

uₙ = a + (n−1)d. Term n from term 1 takes (n−1) steps, not n.
Sₙ = ½n[2a+(n−1)d] = ½n(a+l), l = a+(n−1)d = last term. The (a+l) form needs no d.
Sum proof: write forwards, write reversed, add — every pair totals 2a+(n−1)d, n pairs.
Arithmetic: uₙ₊₁−uₙ = d, constant. Geometric (later): uₙ₊₁/uₙ = r, constant. Check which before choosing a formula.
Increasing: uₙ₊₁>uₙ ∀n. Decreasing: uₙ₊₁<uₙ ∀n. Periodic, period k: uₙ₊ₖ=uₙ ∀n.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was independently verified against the primary Pearson document, not carried over from prior course material. One question in this lesson is a real, verbatim reproduction of a genuine Pearson question, marked accordingly: the marked-solution's part (a) is January 2020 Q8(i) (WMA12_01, Publications Code `P60571RA`), reproduced with its own real B1/M1/A1* mark scheme, 3 marks as printed on the question paper. Every other question in this lesson — prequestion, worked chain, chain drill, method comparison, marked-solution part (b) and MCQ alike — is VERIDIAN-original wording, inspired by confirmed real question types, never a reproduction of a real Pearson question; and because those questions are original, their per-line mark allocations are modelled on verified mark-scheme conventions (what M, A and B marks mean, when follow-through and cso/ag apply, which routes earn the method mark for a three-term quadratic) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 12 marks

An arithmetic sequence has $u_1 = 6$ and $u_4 = 18$ . What is the common difference $d$ ?

$4$
Correct. $u_4 = a + 3d$ , so $6 + 3d = 18$ , giving $3d = 12$ and $d = 4$ . Check: $u_1=6, u_2=10, u_3=14, u_4=18$ — three steps of $+4$ from $u_1$ to $u_4$ , matching $(4-1)=3$ applications of the common difference.
B $12$
This reports $u_4 - u_1 = 12$ directly, as if that difference were $d$ itself, rather than $3d$ — the total change across three steps, not the size of one step. Dividing by the number of steps (3, not 4) is the part that got skipped.
C $3$
This divides the total change (12) by 4 instead of 3 — using n itself instead of $(n-1)$ as the number of steps between $u_1$ and $u_4$ . There are only three steps between the 1st and 4th terms, not four.
D $4.5$
This divides $u_4$ itself (18) by 4, rather than dividing the change between $u_1$ and $u_4$ by the number of steps. $d$ is a rate of change between consecutive terms, not a fraction of any single term's value.

Traps tested: Total change reported as single step · N minus one treated as n · Term value divided instead of difference

Question 22 marks

A sequence is defined by $u_1 = 3$ and $u_{n+1} = 2u_n - 1$ . What is the value of $u_3$ ?

$9$
Correct. $u_2 = 2(3) - 1 = 5$ , then $u_3 = 2(5) - 1 = 9$ . Both steps of the rule — double, then subtract 1 — have to be carried out in full each time; neither part of the rule is optional.
B $11$
This drops the " $-1$ " on the first step only: $u_2 = 2(3) = 6$ , then $u_3 = 2(6) - 1 = 11$ . The recurrence has two operations chained together every single time it is applied, not just after the first term.
C $5$
This is $u_2$ , not $u_3$ — one application of the rule short. Reaching $u_3$ from $u_1$ needs the rule applied twice, and this stops after the first.
D $7$
This treats the sequence as if it were arithmetic, computing $u_1 + 2 \times \text{(some assumed constant difference)}$ . There is no constant difference here — $u_2-u_1 = 2$ but $u_3-u_2=4$ — because this recurrence is not of the arithmetic form $u_{n+1}=u_n+d$ at all; it depends on $u_n$ itself, not just on a fixed step.

Traps tested: Recurrence operation dropped on first step · Recurrence applied one time too few · Non arithmetic recurrence treated as arithmetic

Question 31 mark

The sequence defined by $u_n = (-1)^n$ is . What is its period?

$2$
Correct. $u_1=-1, u_2=1, u_3=-1, u_4=1, \ldots$ — the pattern $-1, 1$ repeats every 2 terms, and $u_{n+2} = u_n$ for every $n$ : $(-1)^{n+2} = (-1)^n(-1)^2 = (-1)^n$ .
B $1$
Period 1 would mean every term is identical to the one before it ( $u_{n+1}=u_n$ for all n) — a constant sequence. This sequence alternates between two different values, so consecutive terms are never equal; the repeat only reappears every second term.
C $4$
$u_{n+4}=u_n$ is true here, but 4 is not the smallest gap at which the sequence repeats — it repeats after just 2 terms already, and repeating after 4 is simply a consequence of repeating after 2 (twice over). The period is the smallest such gap.
DIt is not periodic, since the terms keep changing
A sequence can be periodic while its individual terms are not constant — periodicity means the pattern of values repeats exactly, not that the sequence stops changing. $u_n=(-1)^n$ changes value every single term and is still periodic, with period 2.

Traps tested: Alternating mistaken for period one · Non minimal repeat length given · Periodicity confused with constancy

Question 43 marks

An arithmetic series has first term $10$ and common difference $-2$ . Find $S_{25}$ .

$-350$
Correct. $S_{25} = \frac{1}{2}(25)[2(10)+24(-2)] = 12.5[20-48] = 12.5(-28) = -350$ . With a negative common difference, later terms in the series are negative and large enough in size to pull the total below zero — a negative sum here is a correct result, not a warning sign.
B $-375$
This uses $(n)$ instead of $(n-1)$ inside the bracket: $\frac{1}{2}(25)[20+25(-2)] = 12.5(-30) = -375$ . The bracket needs the number of steps between the 1st and 25th terms, which is 24, not 25.
C $350$
This reaches the correct size but drops the negative sign somewhere in the working — likely by mishandling $24 \times (-2) = -48$ as $+48$ . With $d$ negative, $-4ac$ -style sign tracking is exactly where this question type is designed to catch a slip.
D $248$
This computes $25a + d = 25(10) + (-2) = 248$ , which is not the sum formula at all — it multiplies the first term by the number of terms and then just adds $d$ once, rather than accounting for every term's own individual shift from $a$ .

Traps tested: N minus one treated as n · Negative common difference sign lost · Sum formula replaced with wrong expression

Question 52 marks

In the standard proof that $S_n = \frac{1}{2}n[2a+(n-1)d]$ , why is the series written a second time in reverse order before the two versions are added?

So that every matched pair of terms — one from each version — sums to the same constant value, $2a+(n-1)d$ , letting the whole total be written as $n$ copies of that one number
Correct. Reversing means that as one version steps up through the sequence, the other steps down by the same amount at the same time, so every pairing lands on the identical total. That is the entire mechanism: without it, there is no reason the $n$ terms being added would combine into anything simpler than they already were.
BSo that the negative terms cancel out
An arithmetic series does not need to contain any negative terms at all for this proof to work — it holds identically for a series of all-positive terms. The proof is about the pairs summing to a constant, not about cancellation of sign.
CBecause the formula only works when $n$ is even
The formula $S_n = \frac{1}{2}n[2a+(n-1)d]$ holds for every positive integer $n$ , odd or even — try it on a 3-term or 5-term series and it still gives the correct total. The proof's pairing argument works regardless of parity; it does not require the terms to split into complete pairs with nothing left over.
DTo turn the arithmetic series into a geometric one, which is easier to sum
Reversing the order of an arithmetic series produces another arithmetic series (with common difference $-d$ instead of $d$ ) — it stays arithmetic throughout. Nothing about this proof involves a geometric series at any stage.

Traps tested: Reversal purpose confused with cancellation · Sum formula wrongly restricted to even n · Arithmetic and geometric formulas confused

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Geometric Series and the Sum to Infinity

An arithmetic sequence moves by adding the same amount every step; a geometric one moves by multiplying by the same amount every step. That single swap — +d becomes \times r — changes everything downstream, and changes it in ways that are derivable rather than arbitrary: the nth-term formula becomes u_n = ar^{n-1}, the proof of the sum formula stops working by reversing-and-adding and starts working by multiplying-and-subtracting, and something genuinely new appears that has no arithmetic counterpart at all. Add infinitely many terms of an arithmetic series and the total always runs away. Add infinitely many terms of a geometric series and, provided |r| < 1, it settles on a finite number — S_\infty = \frac{a}{1-r} — because each term is a fixed fraction of the one before it and the terms shrink toward nothing fast enough to leave a finite total behind. The specification asks for the proof of the finite sum formula by name, and examiners have said outright that students should know the standard proofs for both the arithmetic and the geometric series. This lesson derives both formulae, and derives the |r| < 1 condition rather than quoting it.

55 min

Sequences, Recurrence Relations and Arithmetic Series

Two ways to describe a sequence — a formula, and a rule for the next term

Arithmetic sequences, and three families of behaviour

Why un=a+(n−1)du_n = a + (n-1)dun​=a+(n−1)d — it falls out of the recurrence, it is not a separate rule to learn

Why the sum formula works — a symmetry, not a fact to memorise

Two given terms, secretly a simultaneous-equations question — finding aaa, ddd, and then S20S_{20}S20​

Complete the chain — find aaa, ddd and S15S_{15}S15​ given u3=20u_3 = 20u3​=20 and u8=50u_8 = 50u8​=50

Deep dive

Practice this for real

Why $u_n = a + (n-1)d$ — it falls out of the recurrence, it is not a separate rule to learn

Two given terms, secretly a simultaneous-equations question — finding $a$ , $d$ , and then $S_{20}$

Complete the chain — find $a$ , $d$ and $S_{15}$ given $u_3 = 20$ and $u_8 = 50$