Maths › Sequences and series › Arithmetic series

Arithmetic series

Add the same amount each step and you have an arithmetic sequence; add the sequence up and a two-line trick from the proof unit's toolbox collapses the whole sum. The formula is in the booklet, but the exam wants the proof too, and the proof is the best part.

Year 12-13EDEXCEL 9MA0 4.4, 4.6

Builds on Sequences and sigma notation.

IN THIS TOPIC

  • Use the nth-term formula a + (n − 1)d fluently, including to recover a and d from given terms.
  • Prove and use the sum formula, in both its forms.
  • Apply arithmetic series to modelling, saving schemes included.

WHAT YOU PROBABLY THINK

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 its nth term is

un = a + (n − 1)dNOT IN THE BOOKLET — LEARN IT

on the must-learn list, and with the bracket carrying the whole trap in this topic. The 10th term has taken nine steps, not ten, so it is a + 9d, and the opening lie is off by one step every time.

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 3rd to 10th is seven steps: 7d = 35 − 14 = 21, so d = 3.

Then a + 2d = 14 gives a = 8.

u20 = 8 + 19 × 3 = 65.

Differences of terms count steps between them, which turns simultaneous equations into one subtraction. Seven steps, not ten minus three positions of anything else.

The sum, and its proof

The sum of the first n terms has a closed formula, in the booklet in both dressings,

Sn = n2(2a + (n − 1)d) = n2(a + l)IN THE FORMULAE BOOKLET

with l the last term. The specification requires the proof, and it is two lines. Write the sum forwards, write it backwards, and add: every column totals a + l, there are n columns, and the doubled sum is n(a + l). Halve. ∎

The sum 1 to 10 written forwards above itself written backwards: every column adds to 11, ten columns make 110, and halving gives the sum 55110112911381147115611651174118311921110111the same sum, forwards and backwardsten columns of 11 make 110; halve for one copy: 55
FIG. 1The proof for 1 + 2 + … + 10: forwards over backwards, ten columns of 11, so twice the sum is 110 and the sum is 55.

Applied to 1 + 2 + … + n the proof gives the sum of the first n natural numbers, n(n + 1)/2, a result quoted freely once proved.

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 make a built-in check; when both are quick, run both.

YOUR TURN

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.

The half-n outside the bracket keeps the arithmetic small; multiplying out first is where slips creep in.

Series that save money

Arithmetic series model anything growing by equal instalments, and the exam's favourite setting is the saving scheme. The question “when does the total first pass a target” becomes a quadratic inequality in n, solved and then rounded up, because months come in whole numbers.

A savings scheme paying 20 pounds in month one and 5 pounds more each month after: the running total climbs a steepening staircase, passing the 1000 pound target in month 17 at 1020 pounds£1000 targetmonth 17: £1020running total of an arithmetic series
FIG. 2£20 in month one, £5 more each month. The running total is Sₙ, and it first clears £1000 in month 17, at £1020.

TRY IT UNSEEN

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, 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 to show the crossing.

THE EXAM BIT

  • The nth term has taken n − 1 steps; write the bracket before the numbers.
  • Recover d from term differences by counting the steps between the positions.
  • Quote the sum formula, then substitute; and remember the proof, forwards plus backwards, is itself examinable.
  • In target questions solve the inequality exactly, then round n up and verify both neighbouring sums.
  • Sigma limits that start above 1 mean subtracting two sums: the terms from 5 to 20 are 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.

CHECK YOUR PROGRESS

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  • Use the nth-term formula a + (n − 1)d fluently, including to recover a and d from given terms.
  • Prove and use the sum formula, in both its forms.
  • Apply arithmetic series to modelling, saving schemes included.

No animated video for this topic yet; these notes stand alone.