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TMUA Sequences and Series

A sequence is an ordered list of numbers and a series is their sum. The TMUA covers arithmetic and geometric series, and sequences defined by a recurrence.

Paper 1 and Paper 2 topic. Specification codes: MM2.1, MM2.2, MM2.3, M4.18, M4.19. Last checked 12 November 2026.

Arithmetic series

With first term aa and common difference dd: un=a+(n−1)du_n=a+(n-1)d and Sn=n2(2a+(n−1)d)S_n=\tfrac n2\left(2a+(n-1)d\right).

Geometric series

With first term aa and common ratio rr: un=arn−1u_n=ar^{n-1} and Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r}. If ∣r∣<1|r|<1 the sum to infinity is a1−r\dfrac{a}{1-r}.

Recurrence relations

For un+1=f(un)u_{n+1}=f(u_n), a limit LL (if it exists) must satisfy L=f(L)L=f(L).

Spotting the type

Check differences first. A constant difference means arithmetic, a constant ratio means geometric, and constant second differences mean a quadratic sequence.

The sum of the first nn natural numbers is n(n+1)2\dfrac{n(n+1)}2. A geometric series with ∣r∣≥1|r|\ge1 does not converge, so never use a1−r\dfrac a{1-r} there. Read the first few terms before choosing a formula.

For an arithmetic series, writing the sum forwards and backwards gives the pairing idea behind the formula: every pair of terms from the two ends has the same total.

Check small cases: find the first two or three terms from your formula and compare them with the information in the question before you move on.

Worked examples

Example 1

An arithmetic sequence has third term 11 and eighth term 31. Find the sum of the first 20 terms.

Show the worked solution

5d=31−11=205d=31-11=20, so d=4d=4 and a=11−8=3a=11-8=3. Then S20=10(6+76)=820S_{20}=10(6+76)=820.

Answer: 820

Example 2

A geometric series has first term 12 and sum to infinity 18. Find the sum of the first three terms.

Show the worked solution

121−r=18\dfrac{12}{1-r}=18, so r=13r=\tfrac13. The terms are 12, 4, 4312,\,4,\,\tfrac43, which sum to 523\tfrac{52}{3}.

Answer: 523\tfrac{52}{3}

Example 3

A sequence has un+1=12un+3u_{n+1}=\tfrac12u_n+3 and settles at a limit LL. Find LL.

Show the worked solution

At the limit L=12L+3L=\tfrac12L+3, so 12L=3\tfrac12L=3 and L=6L=6.

Answer: 6

Common mistakes

  • Using nn instead of n−1n-1 in a+(n−1)da+(n-1)d or arn−1ar^{n-1}.
  • Using the sum to infinity formula when ∣r∣≥1|r|\ge1.
  • Mixing up the number of terms with the position of the last term.
  • Assuming a recurrence has a limit without checking it converges.

Test yourself on this topic with our free practice questions.

Quick answers

What is the formula for the sum of an arithmetic series?

Sn=n2(2a+(n−1)d)S_n=\tfrac n2(2a+(n-1)d), or n2(first+last)\tfrac n2(\text{first}+\text{last}).

When does a geometric series have a sum to infinity?

When ∣r∣<1|r|<1. Then the sum is a1−r\dfrac{a}{1-r}.

How do you find the limit of a recurrence relation?

Replace both un+1u_{n+1} and unu_n with LL and solve the equation.

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

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