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Sum of an Arithmetic Series (Edexcel IGCSE Maths A (Modular))

Revision Note

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Edexcel Maths A (Modular): Higher Unit 2Edexcel Maths A: Higher

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Sum of an Arithmetic Series

What is the sum of an arithmetic sequence?

  • The sum of an arithmetic sequence (a series) means the terms in an arithmetic sequence are added together

    • For example, the sum of the first 5 terms in the sequence 2, 4, 6, 8, 10, 12, ... is

      • 2 + 4 + 6 + 8 + 10 = 30

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

  • The formula for the sum of the first n terms in an arithmetic sequence is

    S subscript n equals n over 2 open square brackets 2 a plus open parentheses n minus 1 close parentheses d close square brackets

    • a is the first term

    • d is the common difference

    • n is the number of terms being added

    • S subscript n is the sum of the first n terms

How do I use the formula for the sum of an arithmetic sequence?

  • You may have to substitute values into the formula to find S subscript n

    • For example, if a equals 2 and d equals 5 then the sum of the first ten terms is S subscript 10

      • Substitute a equals 2, d equals 5 and n equals 10 into the formula

  • You may have to form equations in terms of a and d when given information about the sum of the first n terms

    • This may lead to simultaneous equations

      • The second equation could come from the formula for the nth term, u subscript n equals a plus open parentheses n minus 1 close parentheses d

  • You may have to form a quadratic equation or quadratic inequality in n

    • When solving, remember that n must be a positive integer

Examiner Tips and Tricks

The formula S subscript n equals n over 2 open square brackets 2 a plus open parentheses n minus 1 close parentheses d close square brackets is given in the formula booklet.

Worked Example

An arithmetic sequence is given by 6, 13, 20, 27, ...

(a) Find the sum of the first twenty terms.

The formula for the sum of the first n terms in an arithmetic sequence is given by S subscript n equals n over 2 open square brackets 2 a plus open parentheses n minus 1 close parentheses d close square brackets
Here, a equals 6, d equals 7 and n equals 20, so substitute these in

table row cell S subscript 20 end cell equals cell 20 over 2 open square brackets 2 cross times 6 plus open parentheses 20 minus 1 close parentheses cross times 7 close square brackets end cell row blank equals cell 10 open parentheses 12 plus 19 cross times 7 close parentheses end cell row blank equals 1450 end table

The sum of the first twenty terms is 1450

(b) What is the minimum number of terms required for the sum of the first n terms to exceed 4000?

This question does not tell you the value of n
You know a equals 6 and d equals 7 from before
Use the formula for the sum of the first n terms to write S subscript n greater than 4000

n over 2 open square brackets 2 cross times 6 plus open parentheses n minus 1 close parentheses cross times 7 close square brackets greater than 4000

Multiply both sides by 2 and simplify

table row cell n open square brackets 12 plus 7 open parentheses n minus 1 close parentheses close square brackets end cell greater than 8000 row cell n open square brackets 12 plus 7 n minus 7 close square brackets end cell greater than 8000 row cell n open parentheses 5 plus 7 n close parentheses end cell greater than 8000 row cell 5 n plus 7 n squared end cell greater than 8000 end table

This is a quadratic inequality
Bring all the terms to the left-hand side

7 n squared plus 5 n minus 8000 greater than 0

If you know how to solve a quadratic inequality, you can use that method
If not, make the inequality an equals sign and solve (e.g. the quadratic formula)

n equals fraction numerator negative 5 plus-or-minus square root of 5 squared minus 4 cross times 7 cross times open parentheses negative 8000 close parentheses end root over denominator 14 end fraction

This gives

n equals 33.450913... or n equals negative 34.165199...

n is the number of terms so cannot be negative; it must be near 33
Test n equals 33 to see if S subscript n greater than 4000 from the original question

S subscript 33 equals 33 over 2 open square brackets 2 cross times 6 plus open parentheses 33 minus 1 close parentheses cross times 7 close square brackets equals 3894 less than 4000

This is not enough terms for the sum to exceed 4000
Test n equals 34 to see if S subscript n greater than 4000

S subscript 34 equals 34 over 2 open square brackets 2 cross times 6 plus open parentheses 34 minus 1 close parentheses cross times 7 close square brackets equals 4250 greater than 4000

This is the first time the sum exceeds 4000

You need a minimum of 34 terms for the sum to exceed 4000

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    Author
    Reviewer
    Mark Curtis

    Author: Mark Curtis

    Expertise: Maths

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.