OCR A Level Mathematics A: Sequences, sigma notation, arithmetic and geometric series and modelling (H240) – Revision Notes

These revision notes condense sections 1.04e to 1.04k of the OCR A Level Mathematics A specification (Version 3.1, August 2026, for first assessment in 2018): sequences, sigma notation, arithmetic and geometric series, the sum to infinity and modelling. All of this content is Stage 2, so it is A Level only. For full explanations and fuller worked examples, read the study guide.

Then test yourself with the practice questions. The course page is OCR A Level Mathematics, the printable list is the checklist, and the free diagnostics show where to focus.

Definitions (1.04e, 1.04f)

Term	Meaning
Sequence	An ordered list u₁, u₂, u₃, …
Series	A sum of terms of a sequence
Finite / infinite	The list stops / goes on for ever
Formula for nth term	uₙ = f(n): any term directly
Recurrence	xₙ₊₁ = f(xₙ) plus a starting term: term by term
Increasing	uₙ₊₁ > uₙ for all n
Decreasing	uₙ₊₁ < uₙ for all n
Periodic	Terms repeat in a cycle; order = cycle length
Convergent	Terms (or the sum) approach a limit
Divergent	No limit: grows without bound, or oscillates without settling

Limit of a recurrence: if xₙ → L then L = f(L). Solve this equation. Example: xₙ₊₁ = 0.4xₙ + 9 gives L = 0.4L + 9, so L = 15.

Proving monotonic: show uₙ₊₁ − uₙ has one sign for every n. Spotting periodic: compute until a term repeats the start, then use remainders on division by the order.

Sigma notation (1.04g)

  k
  Σ  f(r)  =  f(m) + f(m+1) + ... + f(k)      (k − m + 1 terms)
 r=m

• Σ c over k − m + 1 terms is c(k − m + 1), not c.
• A sum of two functions splits into two sums, and a constant multiplier comes outside the sigma.
• To change the start, subtract: sum from 6 to 15 = (sum from 1 to 15) − (sum from 1 to 5).
• To write a series in sigma form, find the rth term, then test the first and last terms.

Formulae

Result	Arithmetic	Geometric
Rule	add d each time	multiply by r each time
nth term	uₙ = a + (n − 1)d	uₙ = a rⁿ⁻¹
Sum of n terms	Sₙ = n/2 (2a + (n − 1)d) = n/2 (a + l)	Sₙ = a(1 − rⁿ)/(1 − r) = a(rⁿ − 1)/(r − 1)
Sum to infinity	none (diverges unless all terms 0)	S∞ = a/(1 − r), only if
Finding the parameter	subtract two terms and divide by the gap	divide two terms, take a root

The specification gives a formulae sheet in each paper and lists the two nth term formulae as ones to know. Learn every line above anyway.

Method boxes

Arithmetic problem

1. Write two equations in a and d (from terms or sums).
2. Subtract to find d, then substitute for a.
3. Write uₙ or Sₙ in terms of n.
4. For an inequality in n, solve the quadratic, then test the integers either side.

Geometric problem

1. Divide terms to get rᵏ for a gap of k places.
2. Take the root. If the gap is even, there may be a negative ratio as well as a positive one, so check both against the given information.
3. Find a from one term.
4. State |r| < 1 before using S∞.

Modelling problem

1. Decide: fixed amount (arithmetic) or fixed percentage (geometric).
2. Identify a, d or r, and what n counts. Be exact about whether the first term is the start or the end of the first period.
3. Form the equation or inequality.
4. Solve with logarithms: log(rⁿ) = n log r. Flip the inequality when dividing by a negative number, which happens when r < 1.
5. Give a whole number and check either side, then answer in context with units.

Small worked reminders

Recurrence limit. xₙ₊₁ = 0.4xₙ + 9, x₁ = 5: 5, 11, 13.4, 14.36, … approaching 15.

Sigma. Σ (2r + 3) for r = 1 to 10 = 2(55) + 3(10) = 140, using the arithmetic formula: terms 5 to 23, so 10/2 × (5 + 23) = 140.

Geometric. 2, 6, 18, …: a = 2 and r = 3, so S₅ = 2(3⁵ − 1)/2 = 242.

Infinite. 9 + 9/4 + 9/16 + … = 9/(1 − ¼) = 12.

Alternating. 5/2 − 3/2 + 9/10 − …: r = −3/5, so S∞ = (5/2)/(8/5) = 25/16.

Recurring decimal. 0.1818… = 0.18/0.99 = 2/11.

Doubling inequality. 3 × 1.07ⁿ > 20 gives n > ln(20/3)/ln 1.07 = 28.04…, so n = 29. Check: 3 × 1.07²⁸ ≈ 19.95 and 3 × 1.07²⁹ ≈ 21.34.

Common modelling set-ups

Situation	Model	Ratio or difference
Simple interest of 3% on £2000	arithmetic, d = 60	a = 2000
Compound interest of 3% a year	geometric	r = 1.03
Value falls 8% a year	geometric	r = 0.92
Fixed monthly rise in saving	arithmetic	d = the rise
Regular payment into an account	sum of a geometric series	r = 1 + rate; list the payments one by one first
Repeated dose with a fraction removed	geometric with sum to infinity as the long-run level	r = fraction remaining

For a regular saver, write the value of each payment at the end, starting with the earliest (largest) payment or the latest (smallest). That list is the geometric series, and a and r then come from it.

Must-know distinctions

• Arithmetic or geometric: constant difference or constant ratio? Compare the gaps between neighbouring terms, then compare the ratios.
• Sequence or series: the list or the total. Sₙ is a series; uₙ is a term. Note that uₙ = Sₙ − Sₙ₋₁ for n ≥ 2.
• Convergent or divergent: for a geometric series it depends on |r| < 1 only, not on the size of a.
• Simple or compound interest: simple interest is arithmetic (same amount each year); compound is geometric.
• Increase of 12% or increase to 112%: both give r = 1.12. A decrease of 12% gives r = 0.88.
• Order of a periodic sequence: the number of terms in one cycle. The sequence x, y, x, y, … has order 2.

Quick self-test

1. A sequence has uₙ = 2n² − 11n. Which term is the first to be positive?
2. Find the 4th term of xₙ₊₁ = 3xₙ − 2 with x₁ = 4.
3. Evaluate Σ (3r + 1) from r = 2 to 8.
4. An arithmetic series has first term 7, last term 95 and common difference 4. How many terms does it have, and what is its sum?
5. The 6th term of an arithmetic sequence is 9 and the 11th term is 24. Find a and d.
6. Find the 7th term and the sum of the first 7 terms of the geometric sequence 3, 6, 12, …
7. Find S∞ for 20 + 8 + 3.2 + …
8. Find the range of x for which a geometric series with ratio (x − 4)/3 converges.
9. A sequence has x₁ = 1 and xₙ₊₁ = 4/xₙ. State its order and find x₅₀.
10. A car is worth £16 000 and loses 15% of its value each year. In how many whole years is it first worth less than £6000?
11. A savings plan puts in £200 in year 1 and £15 more each year than the year before. How much is paid in over 12 years?

Answers

1. u₁ = −9, u₂ = −14, u₃ = −15, u₄ = −12, u₅ = −5, u₆ = 6. The first positive term is u₆ (also 2n² − 11n > 0 gives n > 5.5).
2. x₂ = 10, x₃ = 28, x₄ = 82.
3. Terms 7, 10, …, 25 (7 terms): 7/2 × (7 + 25) = 112.
4. 95 = 7 + (n − 1)4 gives n = 23. S₂₃ = 23/2 × (7 + 95) = 1173.
5. 5d = 15, so d = 3 and a = 9 − 15 = −6.
6. u₇ = 3 × 2⁶ = 192. S₇ = 3(2⁷ − 1) = 381.
7. r = 0.4, so S∞ = 20/0.6 = 100/3 (33.3 to 3 s.f.).
8. |x − 4| < 3, so 1 < x < 7.
9. x: 1, 4, 1, 4, … Order 2; x₅₀ is an even term, so x₅₀ = 4.
10. 16 000 × 0.85ⁿ < 6000 gives n > ln(0.375)/ln(0.85) = 6.03…, so 7 years (after 6 years: £6034.39, after 7 years: £5129.23).
11. Arithmetic with a = 200, d = 15: S₁₂ = 12/2 × (400 + 165) = £3390.

Where marks are usually lost

• Writing the answer to a “find n” inequality as a decimal, or not checking the integers either side.
• Quoting S∞ with no check on |r|, or writing r < 1 when r could be below −1.
• Mixing up the number of terms in sigma notation (k − m + 1) and slipping by one.
• Raising a negative ratio to a power without brackets.
• Using 0.15 as the ratio for a 15% fall.
• Failing to give a reason when asked to show a sequence is periodic or increasing. Show the working, not just the first few terms.
• Rounding intermediate values in modelling problems, so the final figure drifts.
• Giving an answer without context (units, years, mg) when the question is set in a real situation.

Go on to the practice questions when you can do the self-test without notes.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Sections 1.04e to 1.04k.