Series for Edexcel IGCSE Further Pure Mathematics: Methods and Worked Examples

Series is one of the most heavily examined areas of edexcel igcse further pure mathematics, and the techniques you learn here connect directly to the binomial section and to calculus

A series is the sum of the terms of a sequence. In many mathematical settings across Europe and beyond, series are encountered early because they bridge arithmetic and algebraic reasoning. In the Pearson Edexcel IGCSE Further Pure Mathematics specification, you need to master sigma notation, arithmetic series and geometric series, including convergence conditions and the sum to infinity. These are practical tools: compound interest, population growth models and many physical systems follow geometric patterns, while evenly spaced payment schedules follow arithmetic ones.

These edexcel igcse further pure mathematics series revision notes cover the full section with worked examples, common pitfalls and self-check questions. The focus is on building the exam fluency that turns knowledge into marks.

Sigma notation

Sigma notation provides a compact way to write the sum of a series. The Greek letter sigma (capital S) tells you to add terms. For example, the sum of the first n natural numbers is written with sigma notation as the sum from r = 1 to n of r.

What you need to know for the exam:

• Read the lower and upper limits: the variable below sigma is the starting index, the number above is the finishing index.
• Substitute each value of the index into the general term and add the results.
• Recognise standard results when they appear inside a sigma expression.

Worked example: evaluating a sigma expression

Evaluate the sum from r = 1 to 5 of (3r + 2).

Substitute r = 1, 2, 3, 4, 5:

(3(1) + 2) + (3(2) + 2) + (3(3) + 2) + (3(4) + 2) + (3(5) + 2) = 5 + 8 + 11 + 14 + 17 = 55.

Alternatively, recognise that 3r + 2 is an arithmetic sequence with first term 5, common difference 3, and 5 terms. Sum = (5/2)(5 + 17) = (5/2)(22) = 55.

Arithmetic series

An arithmetic sequence has a constant common difference d between consecutive terms. The general (nth) term is a_n = a + (n - 1)d, where a is the first term.

The sum of the first n terms of an arithmetic series is:

S_n = (n/2)(2a + (n - 1)d) or equivalently S_n = (n/2)(a + l), where l is the last term.

Worked example: finding the sum of an arithmetic series

Find the sum of the first 20 terms of the arithmetic series 7 + 11 + 15 + 19 + ...

First term a = 7, common difference d = 4, n = 20.

S₂₀ = (20/2)(2(7) + (20 - 1)(4)) = 10(14 + 76) = 10(90) = 900.

Worked example: finding the number of terms

An arithmetic series has first term 3 and common difference 5. The sum of the first n terms is 1275. Find n.

S_n = (n/2)(2(3) + (n - 1)(5)) = (n/2)(6 + 5n - 5) = (n/2)(5n + 1).

Set equal to 1275: n(5n + 1)/2 = 1275, so n(5n + 1) = 2550.

5n² + n - 2550 = 0. Using the quadratic formula: n = (-1 + \u221a(1 + 51000))/10 = (-1 + \u221a51001)/10 = (-1 + 225.836...)/10.

Since n must be a positive integer, n = 22 (check: 22(5(22) + 1)/2 = 22(111)/2 = 1221, which does not equal 1275). Recalculate: try n = 22.5, so check n = 22 and n = 23. n = 22: S = 22(111)/2 = 1221. n = 23: S = 23(116)/2 = 1334. Neither gives 1275 exactly, so re-examine: 5n² + n - 2550 = 0 gives discriminant = 1 + 4(5)(2550) = 51001. \u221a51001 = 225.83... This is not an integer, which suggests a check of the problem setup. With first term 3 and common difference 5: actually n(5n + 1) = 2550. Try n = 22: 22(111) = 2442. Try n = 23: 23(116) = 2668. The sum 1275 does not correspond to an integer n with these parameters. Adjusting: with a = 3, d = 4, S_n = (n/2)(6 + (n-1)4) = (n/2)(4n + 2) = n(2n + 1). Set equal to 1275: n(2n + 1) = 1275. Try n = 25: 25(51) = 1275. Correct. So with d = 4, n = 25.

Geometric series

A geometric sequence has a constant common ratio r between consecutive terms. The general term is a_n = ar^n-1, where a is the first term.

The sum of the first n terms of a geometric series is:

S_n = a(1 - rⁿ) / (1 - r) when r is not equal to 1.

Sum to infinity

When |r| < 1, the terms get progressively smaller and the sum converges to a finite limit:

S_infinity = a / (1 - r).

The condition |r| < 1 is essential and is examined directly. If |r| >= 1, the series diverges and has no finite sum.

Worked example: finding the common ratio and sum to infinity

A geometric series has first term 12 and third term 3. Find the common ratio and the sum to infinity.

Third term: ar² = 3, so 12r² = 3, giving r² = 1/4, so r = 1/2 or r = -1/2.

Both values satisfy |r| < 1, so both give a convergent series.

If r = 1/2: S_infinity = 12 / (1 - 1/2) = 12 / (1/2) = 24.

If r = -1/2: S_infinity = 12 / (1 - (-1/2)) = 12 / (3/2) = 8.

Worked example: finding n given a sum

The first term of a geometric series is 5 and the common ratio is 2. Find the smallest value of n for which S_n > 10000.

S_n = 5(2ⁿ - 1) / (2 - 1) = 5(2ⁿ - 1).

Set 5(2ⁿ - 1) > 10000: 2ⁿ - 1 > 2000, so 2ⁿ > 2001.

Take log₁₀: n log₁₀ 2 > log₁₀ 2001, so n > log₁₀ 2001 / log₁₀ 2 = 3.3013... / 0.3010... = 10.97.

Since n must be an integer, n = 11. Check: S₁₁ = 5(2¹¹ - 1) = 5(2047) = 10235 > 10000. Confirmed.

Worked example: recurring decimal as a geometric series

Express 0.272727... as a fraction.

0.272727... = 27/100 + 27/10000 + 27/1000000 + ...

This is a geometric series with a = 27/100 and r = 1/100.

S_infinity = (27/100) / (1 - 1/100) = (27/100) / (99/100) = 27/99 = 3/11.

Practice questions

Test your command of series edexcel igcse with these edexcel igcse further pure mathematics practice questions.

1. The 5th term of an arithmetic series is 23 and the 12th term is 51. Find the first term and the common difference.
2. A geometric series has first term 8 and common ratio 3/4. Find S₁₀ to 3 significant figures and S_infinity.
3. The sum of the first n terms of an arithmetic series is S_n = 3n² + 2n. Find the 10th term.
4. Express the sum from r = 1 to 20 of (4r - 1) as a single number.
5. A geometric series has sum to infinity 40 and first term 10. Find the common ratio and the sum of the first 5 terms.

Solutions

Question 1: a + 4d = 23 and a + 11d = 51. Subtracting: 7d = 28, so d = 4. Then a = 23 - 16 = 7.

Question 2: S₁₀ = 8(1 - (3/4)¹⁰) / (1 - 3/4) = 8(1 - 0.05631...) / (1/4) = 32(0.94369...) = 30.2 (3 s.f.). S_infinity = 8 / (1 - 3/4) = 8 / (1/4) = 32.

Question 3: The nth term for n >= 2 is S_n - S_n-1 = (3n² + 2n) - (3(n-1)² + 2(n-1)) = 3n² + 2n - 3n² + 6n - 3 - 2n + 2 = 6n - 1. The 10th term = 6(10) - 1 = 59.

Question 4: This is an arithmetic series with first term 3 (when r = 1), last term 79 (when r = 20), and 20 terms. Sum = (20/2)(3 + 79) = 10(82) = 820.

Question 5: S_infinity = a / (1 - r), so 40 = 10 / (1 - r), giving 1 - r = 1/4, so r = 3/4. S₅ = 10(1 - (3/4)⁵) / (1/4) = 40(1 - 243/1024) = 40(781/1024) = 31240/1024 = 30.5078... = 30.5 (3 s.f.).

The igcse 4PM1 series section rewards students who keep their algebraic working tidy and who verify each answer by substitution. These edexcel igcse further pure mathematics notes should serve as your reference sheet for the key formulae and standard methods. For additional edexcel igcse further pure mathematics explained content and more practice, use edexcel igcse further pure mathematics revision notes on the Green Bridge CBT platform to consolidate your understanding of arithmetic and geometric series across all contexts the exam tests.