Maths › Sequences and series › Arithmetic series
Arithmetic series
Add the same amount at every step and you have an arithmetic sequence. Add the sequence up and a two-line trick collapses the whole thing. The sum formula is printed in the booklet, but the specification also wants you to be able to prove it, and the proof is the best part of the topic.
Builds on Sequences and sigma notation.
IN THIS TOPIC
- Use un = a + (n − 1)d fluently, including to recover a and d from two given terms.
- Prove the sum formula, and use it in both of its forms.
- Apply arithmetic series to modelling, saving schemes included.
COMMON MISCONCEPTION
The 10th term of an arithmetic sequence is a + 10d.
The nth term
An arithmetic sequence climbs by a fixed common difference d from a first term a, so the nth term is
which is on the must-learn list. The bracket holds the one real trap of the topic. Reaching the 10th term takes nine steps, not ten, so it is a + 9d; a + 10d lands one step too far along.
WORKED EXAMPLE
Two terms pin the sequence
The 3rd term of an arithmetic sequence is 14 and the 10th term is 35. Find a and d, and the 20th term.
From the 3rd to the 10th is seven steps, so 7d = 35 − 14 = 21 and d = 3.
Then a + 2d = 14 gives a = 8.
u20 = 8 + 19 × 3 = 65.
Subtracting the two given terms counts the steps between them and eliminates d in one step. Seven steps. Not ten minus three of anything else.
The sum, and its proof
The sum of the first n terms closes into a formula, printed in the booklet in two dressings,
with l the last term. The specification wants the proof known, and the proof is two lines. Write the sum forwards. Write it again backwards underneath. Add. Every column totals a + l and there are n columns, so twice the sum is n(a + l), and halving finishes it. ∎
Run the same argument on 1 + 2 + … + n and you get the sum of the first n natural numbers, n(n + 1)/2. Once proved, quote it freely.
WORKED EXAMPLE
A sum from scratch
Find the sum of the first 20 terms of 5 + 9 + 13 + …
Here a = 5 and d = 4, so S20 = 10 × (10 + 19 × 4) = 10 × 86 = 860.
The last-term form agrees. l = 5 + 19 × 4 = 81, and 10 × (5 + 81) = 860.
Two forms of one formula give you a built-in check. When both are quick, run both.
GUIDED PRACTICE
Recover, then sum
Using a = 8 and d = 3 from the last worked example's sequence, find S15, before opening the working.
Show the working
S15 = (15/2)(2 × 8 + 14 × 3) = 7.5 × 58 = 435.
Leave the half-n outside the bracket while you work. Multiplying it through first is where the slips creep in.
Series that save money
Arithmetic series model anything that grows by equal instalments, and the examiner's favourite setting by some distance is the saving scheme. Asking when the total first passes a target turns into a quadratic inequality in n, which you solve and then round up, because months arrive whole.
INDEPENDENT PRACTICE
When does the pot pass £1000?
A saver deposits £20 in month 1, and each month deposits £5 more than the month before. Show that the total after n months is n(5n + 35)/2, and find the month in which the total first exceeds £1000.
Show the working
The deposits are arithmetic with a = 20 and d = 5, so Sn = (n/2)(40 + 5(n − 1)) = n(5n + 35)/2.
Setting n(5n + 35)/2 > 1000 gives n2 + 7n − 400 > 0, and the positive root of the equation is n = (−7 + √1649)/2 = 16.8.
The first whole month past that is month 17, where the total is £1020. Month 16 gives £920 and falls short.
The rounding direction is part of the answer. Round up for “first exceeds”, and quote both neighbouring totals so the examiner can see the crossing.
ASSESSMENT FOCUS
- Reaching the nth term takes n − 1 steps. Write the bracket down before you write any numbers into it.
- Recover d by subtracting two given terms and dividing by the number of steps between their positions.
- Quote the sum formula, then substitute. The forwards-plus-backwards proof is examinable on its own, so rehearse it.
- In target questions, solve the inequality properly, round n up, and verify the sums either side of the crossing.
- Sigma limits starting above 1 mean subtracting two sums. The terms from 5 to 20 give S20 − S4.
CHECK YOURSELF
Prove that the sum of the first n natural numbers is n(n + 1)/2, and evaluate the sum of the first 100.
Show a hint
Forwards, backwards, add the columns.
Show the answer
Write S = 1 + 2 + … + n over S = n + … + 2 + 1 and add. Each of the n columns totals n + 1, so 2S = n(n + 1) and S = n(n + 1)/2. ∎
For n = 100 the formula gives 100 × 101/2 = 5050.
One hundred columns of 101, halved. The proof and the calculation are the same picture.
The nth term is a plus n minus 1 steps of d; count steps, not positions.
Forwards plus backwards makes n columns of a + l; halve for the sum.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the arithmetic series questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Use un = a + (n − 1)d fluently, including to recover a and d from two given terms.
- Prove the sum formula, and use it in both of its forms.
- Apply arithmetic series to modelling, saving schemes included.
Open the full revision checklist to see every objective in the course in one place.