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

This study guide follows the OCR A Level Mathematics A specification (Version 3.1, August 2026, for first assessment in 2018). It covers sections 1.04e to 1.04k of the Pure Mathematics content: sequences, sigma notation, arithmetic and geometric series, the sum to infinity and modelling. The specification marks all of these as Stage 2, so they are A Level only and are not part of the AS qualification. Pure Mathematics content is assessed in all three written papers.

Use it with the revision notes and the practice questions. The course page is OCR A Level Mathematics and the printable list is the checklist. To find your weak spots first, try the free diagnostics. The binomial expansion is a separate topic and is not covered here.

What you must be able to do

Section	Outcome	Stage
1.04e	Work with sequences given by a formula for the nth term or by xₙ₊₁ = f(xₙ): generate terms, find nth terms, comment on behaviour	A Level only
1.04f	Increasing, decreasing and periodic sequences; sequence versus series; finite versus infinite	A Level only
1.04g	Use sigma notation for sums of series	A Level only
1.04h	Arithmetic sequences and series: nth term and Sₙ	A Level only
1.04i	Geometric sequences and series: nth term, finite sum, convergent versus divergent	A Level only
1.04j	Sum to infinity of a convergent geometric series, with	r
1.04k	Modelling with sequences and series, including inequalities with exponentials and logarithms	A Level only

Sequences (1.04e, 1.04f)

A sequence is an ordered list of numbers u₁, u₂, u₃, … A series is the sum of terms of a sequence. A sequence is finite if it stops and infinite if it goes on for ever.

Two ways to define one:

• Formula for the nth term: uₙ = f(n). You can jump straight to any term.
• Recurrence relation: xₙ₊₁ = f(xₙ) with a starting value. You must work term by term.

A sequence is increasing if every term is larger than the one before (uₙ₊₁ > uₙ for all n), decreasing if every term is smaller, and periodic if the terms repeat in a cycle. The order of a periodic sequence is the length of the cycle.

Worked example 1: a formula

uₙ = (3n − 1)/(n + 2). Describe the behaviour.

u₁ = 2/3, u₂ = 5/4, u₃ = 8/5, u₄ = 11/6
uₙ = 3 − 7/(n + 2)

As n grows, 7/(n + 2) falls towards 0 and is positive, so uₙ rises and stays below 3. To prove it is increasing, uₙ₊₁ − uₙ = 7/(n + 2) − 7/(n + 3) = 7/[(n + 2)(n + 3)] > 0. The sequence is increasing and its terms approach 3.

Worked example 2: a periodic recurrence

xₙ₊₁ = 1/(1 − xₙ), x₁ = 3.

x₂ = 1/(1 − 3) = −1/2
x₃ = 1/(1 + 1/2) = 2/3
x₄ = 1/(1 − 2/3) = 3

x₄ = x₁, and each term depends only on the one before, so the cycle repeats: the sequence is periodic with order 3. Then x₉₉ = x₃ = 2/3, because 99 is a multiple of 3.

Worked example 3: a recurrence that settles

xₙ₊₁ = ½xₙ + 6, x₁ = 2 gives 2, 7, 9.5, 10.75, 11.375, … The terms increase and approach a limit L. If xₙ settles, xₙ₊₁ and xₙ both tend to L, so L = ½L + 6 and L = 12.

Compare xₙ₊₁ = 2xₙ − 3, x₁ = 5: 5, 7, 11, 19, 35, … The terms increase without limit, so the sequence diverges.

Sigma notation (1.04g)

Σ means “add up”. The expression below reads: add the values of f(r) for r = m, m + 1, …, k.

  k
  Σ  f(r)
 r=m

The number of terms is k − m + 1. Useful rules: Σ c = c × (number of terms) for a constant c, and Σ [f(r) + g(r)] splits into two sums.

Worked example 4

Evaluate Σ (5r − 3) from r = 4 to 12, and write 3 + 7 + 11 + … + 27 in sigma notation.

Terms: 17, 22, 27, ..., 57 (9 terms, arithmetic)
Sum = 9/2 × (17 + 57) = 333
3 + 7 + 11 + ... + 27 = Σ (4r − 1), r = 1 to 7

Check the second: 4(1) − 1 = 3 and 4(7) − 1 = 27. Different starting values of r give different but equal answers, so always test both ends.

Arithmetic sequences and series (1.04h)

An arithmetic sequence adds a constant common difference d each time. The first term is a and the last term is l.

uₙ = a + (n − 1)d
Sₙ = n/2 × (2a + (n − 1)d) = n/2 × (a + l)

Use the second form of Sₙ when you know the last term. The specification says a formulae sheet is given in each question paper, and its list of formulae to be known includes the nth term of each progression. Learn all of the formulae here regardless.

Worked example 5

An arithmetic sequence has u₆ = −4 and u₁₅ = 32. Find the smallest n for which Sₙ > 500.

a + 5d = −4
a + 14d = 32
Subtract: 9d = 36, so d = 4 and a = −24
Sₙ = n/2 × (−48 + 4(n − 1)) = n(2n − 26) = 2n² − 26n
2n² − 26n > 500
n² − 13n − 250 > 0
Roots: n = (13 ± √1169)/2, so n > 23.6 (the other root is negative)

Test: S₂₃ = 460 and S₂₄ = 528. The smallest n is 24. Always finish with a check on the integers either side.

Worked example 6

Find the sum of the multiples of 7 between 100 and 500.

The first is 105 and the last is 497. Then 497 = 105 + (n − 1)7 gives n = 57. S₅₇ = 57/2 × (105 + 497) = 17 157.

Geometric sequences and series (1.04i)

A geometric sequence multiplies by a constant common ratio r each time.

uₙ = a rⁿ⁻¹
Sₙ = a(1 − rⁿ)/(1 − r) = a(rⁿ − 1)/(r − 1)   (r ≠ 1)

Use whichever form has the positive denominator. If |r| > 1 the terms grow in size and the series is divergent. If |r| < 1 the terms get closer to 0 and the series is convergent. If r is negative the signs alternate.

Worked example 7

A geometric sequence has u₃ = 18 and u₆ = 486. Find a, r and S₁₀.

u₆ / u₃ = r³ = 486/18 = 27, so r = 3
a × 3² = 18, so a = 2
S₁₀ = 2(3¹⁰ − 1)/(3 − 1) = 3¹⁰ − 1 = 59 048

Worked example 8

Find S₆ for the series 40 − 20 + 10 − …

Here a = 40 and r = −½.

S₆ = 40(1 − (−½)⁶)/(1 + ½) = 40 × (63/64) × (2/3) = 105/4

The answer is 26.25. Put brackets round a negative ratio before raising it to a power.

Sum to infinity (1.04j)

If |r| < 1, then rⁿ → 0 as n grows, so Sₙ approaches a limit:

S∞ = a/(1 − r),   provided |r| < 1  (that is, −1 < r < 1)

If |r| ≥ 1 there is no sum to infinity.

Worked example 9

A geometric series has first term 12 and sum to infinity 20. Find r and S₅.

12/(1 − r) = 20, so 1 − r = 3/5 and r = 2/5
S₅ = 12(1 − (2/5)⁵)/(3/5) = 12372/625 = 19.7952

So S₅ ≈ 19.8, just under 20, as expected for a convergent series with positive terms.

Worked example 10

(a) Find the values of x for which 6 + 3/x + 3/(2x²) + … converges, and its sum. (b) Write 0.272727… as a fraction.

(a) Here a = 6 and r = 1/(2x), so |1/(2x)| < 1, which gives |2x| > 1, that is x > ½ or x < −½. Then S∞ = 6/(1 − 1/(2x)) = 12x/(2x − 1).

(b) 0.272727… = 0.27 + 0.0027 + … with a = 0.27 and r = 0.01. So S∞ = 0.27/0.99 = 3/11.

Modelling (1.04k)

Identify what changes by a fixed amount each period (arithmetic: simple interest, a fixed pay rise, a steady training increase) and what changes by a fixed percentage (geometric: compound interest, depreciation, decay). Percentage change of p% per period gives ratio r = 1 + p/100, or 1 − p/100 for a fall.

Worked example 11: an exponential inequality

£8000 earns 3.2% compound interest each year. After how many whole years does it first exceed £10 000?

8000 × 1.032ⁿ > 10 000
1.032ⁿ > 1.25
n ln 1.032 > ln 1.25
n > 7.08...

Check: after 7 years the amount is £9973.51 and after 8 years it is £10 292.66. The answer is 8 years. Dividing by ln 1.032 keeps the inequality direction because ln 1.032 is positive. If the ratio were below 1, the log would be negative and the inequality would reverse.

Worked example 12: a loan

£15 000 is borrowed at 4% a year. A payment of £2400 is made at the end of each year after interest is added. Let Bₙ be the amount owed after n payments.

B₀ = 15 000
Bₙ₊₁ = 1.04Bₙ − 2400
B₁ = 13 200, B₂ = 11 328, B₃ = 9381.12

Solving this recurrence gives Bₙ = 60 000 − 45 000 × 1.04ⁿ. Check n = 1: 60 000 − 46 800 = 13 200. The debt is cleared when 1.04ⁿ ≥ 4/3, so n ≥ 7.33. After 7 payments £783.07 is owed, so an 8th payment is needed (£814.39 settles it).

Worked example 13: a build-up in a body

A patient takes 50 mg of a drug each morning. By the next morning 15% of the drug has gone. Just after dose n, the amount is 50 + 50(0.85) + … + 50(0.85)ⁿ⁻¹.

Just after dose 5: 50(1 − 0.85⁵)/0.15 = 185 mg (3 s.f.). In the long run the amount approaches 50/0.15 = 333 mg (3 s.f.), so it never exceeds this level.

Common errors

• Using n terms when there are n + 1, or the reverse. Count terms with last − first + 1 in sigma notation.
• Using S∞ when |r| ≥ 1, or writing |r| < 1 without stating it.
• Rounding too early in modelling. Keep full values until the last line.
• Reversing the inequality direction wrongly when dividing by a negative logarithm, or forgetting to reverse it.
• Treating a percentage fall as r = 0.15 instead of r = 0.85.
• Giving a non-integer n. Interpret the answer as a whole number of periods and check the neighbours.

Continue with the revision notes and then the practice questions.

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.