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A-Level Sequences and series
What the A-Level syllabus expects for Sequences and series, and how to practise it.
What the syllabus expects
- What sequences and series mean in both finite and infinite cases
- Viewing a sequence as a function y = f(n) with n a positive integer
- How the nth term uₙ connects to Sₙ, the sum of the first n terms
- A sequence defined by a formula for its nth term
- A sequence built from a recurrence uₙ₊₁ = f(uₙ), including generating it with a graphing calculator or computer
- Adding and subtracting two series
- When a series converges and its sum to infinity
- The nth-term and sum formulae for a finite arithmetic series
- The nth-term and sum formulae for a finite geometric series
- The condition under which an infinite geometric series converges
- The sum-to-infinity formula for a convergent geometric series
How it's examined
Questions on this topic most often ask you to find, solve, show, determine. About 10% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
The nth term of sequence V is v_n = a^n / b + b/(a(1 - a^n)) - a/(n + 1), with a, b nonzero reals and a != +-1. For some a, v_n tends to a limit L as n -> infinity. Find, with justification, the values of a for which L exists, and state L in terms of a and b.
Show the worked answer
v_n = a^n/b + b/(a(1-a^n)) - a/(n+1). As n->inf: -a/(n+1)->0 always. If |a|>1: a^n is unbounded, so a^n/b diverges => no limit. If |a|<1 (a nonzero, a!=+-1): a^n->0, so a^n/b->0 and b/(a(1-a^n))->b/(a(1-0))=b/a. All three terms converge. Hence L exists iff |a|<1, i.e. -1<a<1 with a!=0, and then L = b/a.
Example 2 (3 marks)
Priya is preparing for a stair-climbing race by running up the staircase of a tower that has 78 floors. She plans to train on four days each week, covering 9 floors on her very first training day. On every training day after that she climbs 3 more floors than she managed on the previous training day. As soon as she completes 78 floors in a single training day she is deemed ready for the race, and on all later training days she simply keeps climbing 78 floors each day. (c) Work out the total number of floors she has climbed by the end of 20 weeks of training. [3]
Show the worked answer
Daily floors form an AP: 9, 12, 15, ... with a=9, d=3, term n = 3n+6. She reaches 78 when 3n+6=78 => n=24, so on day 24 she climbs exactly 78 (first time). 20 weeks x 4 days = 80 training days. Days 1-24 follow the AP; days 25-80 (56 days) she climbs 78 each. Sum of days 1-24: S24 = 24/2 (9 + 78) = 12 x 87 = 1044. Days 25-80: 56 x 78 = 4368. Total = 1044 + 4368 = 5412.
Example 3 (2 marks)
For the sequence u_1 = k, u_(n+1) = (8 u_n - 14)/(u_n - 1), find the value(s) of k making U a constant sequence.
Show the worked answer
A constant sequence needs u_(n+1)=u_n=k. So k=(8k-14)/(k-1). Multiply: k(k-1)=8k-14 => k²-k=8k-14 => k²-9k+14=0 => (k-2)(k-7)=0. So k=2 or k=7.
More worked questions on this topic
- A sequence has u1=2 and u_{n+1}=1/(1−u_n) for n≥1. Find u2, u3 and u4, and hence give u2025. (3 marks)
- Show that after her nth monthly repayment the outstanding loan is 23760(1.004)ⁿ − 250y(1.004ⁿ − (3 marks)
- A sequence has nth term u_n = a n² + b n + c. Its opening three terms are 2, 6 and 12. Work out (3 marks)
- With the value of r from part (i), find the smallest n for which the sum of all terms beyond th (3 marks)
- A geometric sequence {u_n} has first term a > 0 and common ratio 3/2. Define {v_n} by v_n = 9/u (4 marks)
- Continuing part (a) with r = 0.6639, build a fresh series using every even-indexed term of that (4 marks)
- Let an arithmetic series have common difference d and first term a (both non-zero); a convergen (4 marks)
More A-Level H2 Maths topics
Functions · Graphs and their transformations · Equations and inequalities · Vectors in two and three dimensions: basic properties · Scalar and vector products · Vector geometry in three dimensions · all of A-Level H2 Maths