Arithmetic Series Applications

This lesson focuses on applying arithmetic series to problem-solving and modelling contexts at A-Level (9MA0). You will also learn to prove the sum formula, a skill Edexcel frequently tests in exam questions.

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Proof of the Sum Formula

This is a classic proof that Edexcel can ask you to reproduce.

Prove that S(n) = n/2 x (2a + (n-1)d)

Write the sum forwards: S(n) = a + (a + d) + (a + 2d) + ... + (a + (n-1)d)

Write the sum backwards: S(n) = (a + (n-1)d) + (a + (n-2)d) + ... + (a + d) + a

Add these two expressions term by term. Each pair sums to: a + (a + (n-1)d) = 2a + (n-1)d

There are n such pairs, so: 2 x S(n) = n x (2a + (n-1)d)

Therefore: S(n) = n/2 x (2a + (n-1)d) as required.

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Using S(n) to Find Individual Terms

A key relationship:

u(n) = S(n) - S(n-1) for n >= 2

This allows you to find any term if you know S(n) as a formula.

Worked Example 1

The sum of the first n terms of a series is S(n) = 3n² + 5n. Find the 10th term.

Method 1: u(10) = S(10) - S(9)

S(10) = 3(100) + 5(10) = 300 + 50 = 350 S(9) = 3(81) + 5(9) = 243 + 45 = 288

u(10) = 350 - 288 = 62

Method 2: Find the general term. u(n) = S(n) - S(n-1) = (3n² + 5n) - (3(n-1)² + 5(n-1)) = 3n² + 5n - 3n² + 6n - 3 - 5n + 5 = 6n + 2

u(10) = 6(10) + 2 = 62. Confirmed.

Check: u(1) = S(1) = 3 + 5 = 8. And 6(1) + 2 = 8. Consistent.

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Modelling with Arithmetic Sequences

Worked Example 2

A company produces 250 items in its first week and increases production by 30 items each week.

(a) Find a formula for the number of items produced in week n.

u(n) = 250 + (n - 1)(30) = 220 + 30n

(b) How many items are produced in total during the first 12 weeks?

S(12) = 12/2 x (2(250) + 11(30)) = 6 x (500 + 330) = 6 x 830 = 4980 items

(c) In which week does the total production first exceed 10,000 items?

S(n) > 10000 n/2 x (500 + 30(n-1)) > 10000 n/2 x (470 + 30n) > 10000 n(470 + 30n) > 20000 30n² + 470n - 20000 > 0 3n² + 47n - 2000 > 0

Using the quadratic formula: n = (-47 + sqrt(2209 + 24000)) / 6 = (-47 + sqrt(26209)) / 6

sqrt(26209) ≈ 161.9

n > (-47 + 161.9) / 6 ≈ 19.15

Total production first exceeds 10,000 in week 20.

Worked Example 3

Logs are stacked in rows. The bottom row has 25 logs, the next has 24, then 23, and so on. How many complete rows can be formed if there are 200 logs available?

This is an AP with a = 25, d = -1.

S(n) = n/2 x (2(25) + (n-1)(-1)) = n/2 x (51 - n)

We need S(n) <= 200:

n(51 - n)/2 <= 200 n(51 - n) <= 400 51n - n² <= 400 n² - 51n + 400 >= 0

Roots: n = (51 ± sqrt(2601 - 1600)) / 2 = (51 ± sqrt(1001)) / 2

sqrt(1001) ≈ 31.64

n <= (51 - 31.64)/2 ≈ 9.68 or n >= (51 + 31.64)/2 ≈ 41.32

Since n must be positive and at most 25 (rows cannot have 0 or fewer logs):

n = 9 complete rows can be formed.

Check: S(9) = 9/2 x (51 - 9) = 9/2 x 42 = 189 <= 200 S(10) = 10/2 x (51 - 10) = 5 x 41 = 205 > 200

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Working with Unknown Parameters

Worked Example 4

The first term of an AP is a and the common difference is d. Given that the sum of the first 10 terms is 200 and the sum of the next 10 terms is 600, find a and d.

S(10) = 10/2 x (2a + 9d) = 5(2a + 9d) = 200 So 2a + 9d = 40 ... (1)

Sum of terms 11 to 20 = S(20) - S(10) = 600 So S(20) = 800

S(20) = 20/2 x (2a + 19d) = 10(2a + 19d) = 800 So 2a + 19d = 80 ... (2)

Subtract (1) from (2): 10d = 40, so d = 4

Substitute: 2a + 36 = 40, so a = 2

Answer: a = 2, d = 4

Worked Example 5

Three consecutive terms of an AP are (k + 1), (2k + 3), (4k - 1). Find k.

In an AP, the middle term is the mean of the other two:

2(2k + 3) = (k + 1) + (4k - 1)

4k + 6 = 5k

k = 6

Check: terms are 7, 15, 23 with d = 8.

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Exam-Style Problem Solving

Worked Example 6

The sum of the first n terms of an arithmetic series is S(n) = n/2 x (3n + 7).

(a) Find the first term.

u(1) = S(1) = 1/2 x (3 + 7) = 5

(b) Find the common difference.

u(2) = S(2) - S(1) = 2/2 x (6 + 7) - 5 = 13 - 5 = 8

d = u(2) - u(1) = 8 - 5 = 3

(c) Find u(n) and verify it gives an arithmetic sequence.

u(n) = S(n) - S(n-1) = n(3n + 7)/2 - (n-1)(3n + 4)/2

= [3n² + 7n - (3n² + 4n - 3n - 4)] / 2

= [3n² + 7n - 3n² - n + 4] / 2

= (6n + 4) / 2

= 3n + 2

Check: u(1) = 5, u(2) = 8, u(3) = 11. Common difference = 3.

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Exam Tips

Tip	Detail
Proof questions	Learn the sum formula proof — it appears regularly as a 3-4 mark question
u(n) = S(n) - S(n-1)	This is essential for finding terms from a given sum formula
Check u(1) separately	When finding u(n) from S(n), always verify that u(1) = S(1)
Context problems	Interpret n carefully (week 1 vs week 0, row number, etc.)
Inequality answers	When n must be an integer, state the rounding direction with justification

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Practice Problems

1. The sum of the first n terms of a series is S(n) = 2n² + 3n. Find u(15). [Answer: 61]
2. Prove that S(n) = n/2 x (a + l) where l is the last term.
3. A runner jogs 2 km on day 1 and increases the distance by 0.5 km each day. After how many days has she run a total of 50 km? [Answer: 10 days]
4. The 5th and 12th terms of an AP are 23 and 51 respectively. Find S(20). [Answer: 740]
5. Three terms k, 2k + 1, 11 form an AP. Find k. [Answer: k = 3]

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Summary

• The sum formula S(n) = n/2(2a + (n-1)d) can be proved by writing the sum forwards and backwards.
• u(n) = S(n) - S(n-1) connects individual terms to the sum formula.
• Arithmetic series model situations with constant increases (or decreases).
• Always check context: is the answer a positive integer? Does it make physical sense?

A-Level Deep Dive: Arithmetic Series Applications

Spec mapping

Edexcel 9MA0-01 specification section 4 — Sequences and series, sub-strands 4.2 and 4.3 covers sigma notation. Understand and work with arithmetic sequences and series, including the formulae for the $n$nth term and the sum to $n$n terms. Understand and use the terms 'arithmetic progression' (AP) and 'common difference' (refer to the official specification document for exact wording). Although AP results are derived in 4.2, applying them to real-world contexts is examined heavily in 9MA0-01 modelling questions and again in 9MA0-02 statistics-flavoured contexts. Applications appear most commonly as financial mathematics (salary increases, repayment plans, accumulating savings), arrangement problems (stadium seating, performance schedules) and time-to-target problems (the smallest $n$n for which $u_n$un​ exceeds a threshold). The Edexcel formula booklet does provide the AP $n$nth-term formula $u_n = a + (n - 1)d$un​=a+(n−1)d and the sum formula $S_n = \tfrac{n}{2}[2a + (n - 1)d]$Sn​=2n​[2a+(n−1)d] — but it does not provide guidance on choosing AP versus GP, nor on evaluating modelling assumptions, both of which are tested explicitly.

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Worked example with full mark scheme

Question (8 marks):

Anjali begins a new job on a starting salary of £24,000 in year 1. Her contract guarantees an increase of £800 each subsequent year, so her salary forms an arithmetic progression.

(a) Find Anjali's salary in year 10 of the contract. (2)

(b) Find the total amount Anjali will earn during the first 25 years of the contract. (3)

(c) Determine the first year in which Anjali's salary exceeds £40,000. (2)

(d) State one assumption of this model and comment briefly on its plausibility. (1)

Solution with mark scheme:

(a) Step 1 — identify $a$a and $d$d.

$a = 24000$a=24000, $d = 800$d=800. Year 10 corresponds to $n = 10$n=10.

M1 — correct identification of the first term and common difference, with the right value of $n$n. A common error is to use $n = 9$n=9 on the grounds that "after nine increases", giving an answer one year out.

Step 2 — apply the $n$nth-term formula.

$u_{10} = a + (n - 1)d = 24000 + 9 \cdot 800 = 24000 + 7200 = 31200$u10​=a+(n−1)d=24000+9⋅800=24000+7200=31200

A1 — salary in year 10 is £31,200.

(b) Step 1 — apply the sum formula with $n = 25$n=25.

$S_{25} = \dfrac{25}{2}\bigl[2 \cdot 24000 + 24 \cdot 800\bigr] = \dfrac{25}{2}\bigl[48000 + 19200\bigr] = \dfrac{25}{2} \cdot 67200$S25​=225​[2⋅24000+24⋅800]=225​[48000+19200]=225​⋅67200

M1 — correct substitution into $S_n = \tfrac{n}{2}[2a + (n - 1)d]$Sn​=2n​[2a+(n−1)d] with $n = 25$n=25, $a = 24000$a=24000, $d = 800$d=800. The $(n - 1) = 24$(n−1)=24 step is where off-by-one errors creep in.

M1 — correct evaluation of the bracket: $2a + (n - 1)d = 67200$2a+(n−1)d=67200.

Step 2 — compute.

$S_{25} = 12.5 \cdot 67200 = 840000$S25​=12.5⋅67200=840000

A1 — total earnings over 25 years are £840,000.

(c) Step 1 — set up the inequality.

We need the smallest integer $n$n such that $u_n > 40000$un​>40000, i.e. $24000 + (n - 1) \cdot 800 > 40000$24000+(n−1)⋅800>40000.

M1 — correct inequality with $u_n$un​ in terms of $n$n.

Step 2 — solve for $n$n.

$(n - 1) \cdot 800 > 16000 \implies n - 1 > 20 \implies n > 21$(n−1)⋅800>16000⟹n−1>20⟹n>21

The smallest integer $n$n satisfying $n > 21$n>21 is $n = 22$n=22.

A1 — Anjali's salary first exceeds £40,000 in year 22. Note that $n = 21$n=21 gives $u_{21} = 24000 + 20 \cdot 800 = 40000$u21​=24000+20⋅800=40000, which is equal to £40,000, not strictly greater. The strict inequality matters.

(d) B1 — one valid assumption with brief comment, e.g.: "The model assumes the £800 annual increment is fixed in nominal terms regardless of inflation, performance review or contract renegotiation. In practice, salary increases often track inflation (a percentage rise, suggesting a GP) or are subject to discretionary review, so the AP model is a simplification suited to a fixed contractual arrangement."

Total: 8 marks (M3 A4 B1, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A small theatre is designed so that the front row contains 18 seats and each subsequent row contains 3 more seats than the row in front. The theatre has 24 rows in total.

(a) Find the number of seats in the final (24th) row. (2)

(b) Find the total seating capacity of the theatre. (2)

(c) The theatre management wishes to extend the theatre so that the total capacity exceeds 1,200 seats, keeping the same starting row size and common difference. Determine the smallest number of additional rows required. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1b) — substituting $a = 18$a=18, $d = 3$d=3, $n = 24$n=24 into $u_n = a + (n - 1)d$un​=a+(n−1)d.
• A1 (AO1.1b) — $u_{24} = 18 + 23 \cdot 3 = 87$u24​=18+23⋅3=87 seats.

(b)

• M1 (AO1.1b) — substituting into $S_n = \tfrac{n}{2}[2a + (n - 1)d]$Sn​=2n​[2a+(n−1)d] with $n = 24$n=24.
• A1 (AO1.1b) — $S_{24} = 12 \cdot (36 + 69) = 12 \cdot 105 = 1260$S24​=12⋅(36+69)=12⋅105=1260 seats.

(c)

• M1 (AO3.1a) — setting up the inequality $\tfrac{N}{2}[36 + 3(N - 1)] > 1200$2N​[36+3(N−1)]>1200 for total rows $N$N, modelling-context interpretation.
• A1 (AO3.2a) — solving the resulting quadratic $3N^2 + 33N - 2400 > 0$3N2+33N−2400>0, finding $N > 23.0...$N>23.0..., so the smallest $N$N is 24, meaning 0 additional rows are required because the existing 24-row design already exceeds 1,200 ($S_{24} = 1260$S24​=1260). Acceptable equivalent: candidate explicitly notes that $S_{23} = 1173 < 1200$S23​=1173<1200 but $S_{24} = 1260 > 1200$S24​=1260>1200, and concludes 0 additional rows.

Total: 6 marks split AO1 = 4, AO3 = 2. This is a deliberately AO3-heavy specimen — Edexcel uses applied AP questions to test problem-solving and modelling interpretation, not just procedural fluency. The trap in (c) is failing to compare the threshold against the result of (b); a candidate who launches into the quadratic without checking the existing capacity wastes time and risks a sign error.

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Synoptic links

Connects to:

1. AP $n$nth term and sum formulae: every applied AP question reduces, ultimately, to identifying $a$a and $d$d correctly and substituting into one of the two booklet formulae. The applied "dressing" — salary, savings, seats, performances — is contextual; the underlying mechanics are pure substitution and rearrangement. A* candidates make the formula choice automatic and spend their time on interpretation.

2. Modelling assumption-evaluation (AO3): Edexcel routinely awards a final B1 mark for stating and critiquing a modelling assumption. This skill is also tested in mechanics ("assume the string is light and inextensible"), statistics ("assume the sample is representative") and pure ("assume the function is differentiable on the interval"). The phrasing is portable across topics: name the assumption, comment on its plausibility, suggest a refinement.

3. Financial mathematics — AP versus GP: salary increases of a fixed cash amount are AP; salary increases of a fixed percentage are GP. The same principle applies to savings (regular contributions plus simple interest = AP-flavoured; compound interest = GP). Recognising the structural signal — "fixed amount each year" → AP, "fixed percentage each year" → GP — is the synoptic skill bridging sections 4.3 and 4.4.

4. Statistics — modelling and validation: AS-Level statistics (9MA0-02) treats the same modelling-evaluation question — "is this a good model?" — through correlation coefficients, residuals and goodness-of-fit. The pure-mathematics version asks the same question without the data: given the structural form, when does it break?

5. Time-to-target inequalities: "find the smallest $n$n such that $u_n$un​ exceeds $T$T" is the AP cousin of the GP question "find the smallest $n$n such that $ar^{n-1}$arn−1 exceeds $T$T" (which requires logs). Both reduce to integer-valued solutions of inequalities, and both demand a final integer answer with a check that the boundary case is treated correctly (strict vs non-strict).

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Mark-scheme literacy

Applied AP questions on 9MA0 distribute AO marks more evenly than pure procedural questions:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Substituting into $u_n$un​ and $S_n$Sn​ formulae, evaluating arithmetic correctly, identifying $a$a and $d$d from a worded scenario
AO2 (reasoning / interpretation)	20–30%	Justifying the choice of AP over GP, framing the inequality with the correct strict/non-strict form, presenting answers in the requested units
AO3 (problem-solving / modelling)	20–30%	Translating a worded scenario into mathematical structure, evaluating modelling assumptions, drawing context-appropriate conclusions (e.g. integer year numbers, currency rounding)

Examiner-rewarded phrasing for evaluating modelling assumptions: "The model assumes the increment remains constant in nominal terms; in practice, inflation would erode the real value of the increase, suggesting an inflation-adjusted refinement"; "the model treats the contract as continuous over the full period and ignores the possibility of redundancy or promotion, which would change the common difference"; "the AP form is a simplification appropriate for a fixed contractual structure but would be inadequate for performance-linked pay". Phrases that lose marks: "the model is right" or "the model is wrong" without justification; vague claims like "real life is more complicated" with no specific mechanism; commentary on irrelevant aspects of the scenario.

A specific Edexcel pattern to watch: when a question demands a final integer (year number, row count, number of performances), the answer must be presented as a positive integer with units. Decimal answers like "year 21.5" earn no accuracy mark — the candidate must explicitly take the ceiling and justify it ("$n$n must be a positive integer, so the smallest valid $n$n is 22").

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