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Geometric Sequences & Series (Edexcel IGCSE Further Pure Maths)

Revision Note

Amber

Written by: Amber

Reviewed by: Dan Finlay

Geometric Sequences

What is a geometric sequence?

  • In a geometric sequence, there is a common ratio between consecutive terms in the sequence

    • This means each term is multiplied by a common ratio to get the next term

  • The first term of the sequence is denoted by a

  • The common ratio is denoted by r

    • For example, 2, 6, 18, 54, 162, … is a sequence with the rule ‘start at two and multiply each number by three’

      • The first term, bold italic a, is 2

      • The common ratio, bold italic r, is 3

  • A geometric sequence can be

    • increasing (r > 1), or

    • decreasing (0 < r < 1)

  •  If the common ratio is a negative number the terms will alternate between positive and negative values

    • For example, 1, -4, 16, -64, 256, … is a sequence with the rule ‘start at one and multiply each number by negative four’

      • The first term, bold italic a, is 1

      • The common ratio, bold italic r, is -4

  • Terms in a geometric sequence can be referred to

    • by the letter u with

    • a subscript corresponding to its place in the sequence

      • e.g.  u subscript 1 equals a is the first term, u subscript 9 is the ninth term, u subscript n is the nth term, etc.

Geometric Series

What is a geometric series?

  • When the terms of a geometric sequence are added together, that is known as a geometric series

    • The terms (1st term, 2nd term, 3rd term, etc.) are exactly the same in the sequence and series

    • But with the series we're most interested in what happens when the terms are added together

How do I find a term in a geometric series?

  • The bold italic n to the power of bold t bold h end exponent term formula for a geometric series is

u subscript n equals a r to the power of n minus 1 end exponent

  • Where a is the first term, and r is the common ratio

  • This is not given on the exam formula sheet, so make sure you know it

  • The formula allows you to find any term in the geometric series

    • Enter the values of a, r and n and calculate the value of u subscript n

  • Sometimes you will be given a term and asked to find bold italic a or bold italic r

    • Substitute the information you have into the formula and solve the equation

  • Sometimes you will be given two or more consecutive terms and asked to find both bold italic a and bold italic r

    • Find the common ratio by dividing a term by the one immediately before it

      • r equals u subscript n plus 1 end subscript over u subscript n

    • Substitute this and one of the terms into the formula to find the first term

  • Sometimes you may be given a term along with bold italic a and bold italic r and asked to find the value of bold italic n

    • You can solve this using logarithms on your calculator 

How do I find the sum of a geometric series?

  • A geometric series is the sum of the terms in a geometric sequence

    • For the geometric sequence 2, 6, 18, 54, … the geometric series is 2 + 6 + 18 + 54 + …

  • Use the following formula to find the sum of the first n terms of a geometric series:

S subscript n equals fraction numerator a left parenthesis 1 minus r to the power of n right parenthesis over denominator 1 minus r end fraction

  • a is the first term

  • r is the common ratio

  • The formula is given on the exam formula sheet

    • So you don't need to remember it

    • But you do need to know how to use it!

  • If bold italic r bold greater than bold 1 the following rearrangement of the formula might be more convenient:

S subscript n equals fraction numerator a left parenthesis r to the power of n minus 1 right parenthesis over denominator r minus 1 end fraction

  • This version is not on the formula sheet

  • The formula sheet version will always work as well

  • A question will often give you the sum of a certain number of terms and ask you to find the values of bold italic a, bold italic r or bold italic n

    • Substitute the information you have into the formula and solve the equation

Examiner Tips and Tricks

  • The formula for the sum of a geometric series is on the exam formula sheet

    • But the nth term formula is not on the formula sheet

  • You will sometimes need to use logarithms to answer geometric series questions 

    • Make sure you are confident doing this

    • And know how your calculator handles logarithms

Worked Example

The sixth term of a geometric series is 486 and the seventh term is 1458. 

Find

(a) the common ratio of the series

Find the common ratio by dividing a term by the term immediately before it
Here r equals fraction numerator 7 to the power of th space term over denominator 6 to the power of th space term end fraction

r equals 1458 over 486

bold italic r bold equals bold 3

(b) the first term of the series.

Use the nth term formula  u subscript n equals a r to the power of n minus 1 end exponent

For the 6th term, we know  u subscript 6 equals 486,  r equals 3  and  n equals 6

table row 486 equals cell a cross times 3 to the power of open parentheses 6 minus 1 close parentheses end exponent end cell row 486 equals cell a cross times 3 to the power of 5 end cell row 486 equals cell 243 a end cell row a equals cell 486 over 243 end cell end table

bold italic a bold equals bold 2

Worked Example

The first term of a geometric series is 25, and the common ratio is 0.8.

Find the value of the fifth term, as well as the sum of the first 5 terms.

To find the fifth term, use the nth term formula  u subscript n equals a r to the power of n minus 1 end exponent
Here  n equals 5,  a equals 25  and  r equals 0.8

table row cell fifth space term end cell equals cell 25 cross times 0.8 to the power of open parentheses 5 minus 1 close parentheses end exponent end cell row blank equals cell 25 cross times 0.8 to the power of 4 end cell end table

bold fifth bold space bold term bold equals bold 10 bold. bold 24

To find the sum of the first 5 terms, use the sum of a geometric series formula  S subscript n equals fraction numerator a left parenthesis 1 minus r to the power of n right parenthesis over denominator 1 minus r end fraction
Again,  a equals 25  and  r equals 0.8

S subscript 5 equals fraction numerator 25 open parentheses 1 minus 0.8 to the power of 5 close parentheses over denominator 1 minus 0.8 end fraction

Use your calculator to work this out  

bold italic S subscript bold 5 bold equals bold 84 bold. bold 04

Sum to Infinity

What is the sum to infinity of a geometric series?

  • The sum to infinity is the sum of all the terms in a geometric series

    • u subscript 1 plus u subscript 2 plus u subscript 3 plus...  'all the way to infinity'

  • As bold italic n increases the terms of a geometric series may

    • move further away from zero

      • if  r greater than 1  or  r less than negative 1

    • stay the same distance away from zero

      • if  r equals 1  or  r equals negative 1

    • get closer and closer to zero

      • if  negative 1 less than r less than 1

  • If the terms are getting closer to zero then the series is said to converge

  • This means that the sum of the series will approach a finite 'limiting value'

  • As n increases, the sum of the terms will get closer and closer to the limiting value

  • The limiting value is the sum to infinity of the series

    • It is denoted by S subscript infinity

How do I calculate the sum to infinity?

  • First you need to consider the value of bold italic r

    • If  open vertical bar r close vertical bar less than space 1  then the sequence converges

      • open vertical bar r close vertical bar less than 1  is the same as   negative 1 less than r less than 1

      • In this case the sum to infinity can be calculated

    • If  open vertical bar r close vertical bar greater or equal than 1  then the sequence does not converge

      • open vertical bar r close vertical bar greater or equal than 1  is the same as  r less or equal than negative 1  or  r greater or equal than 1

      • In this case the sum to infinity cannot be calculated

  • If vertical line r vertical line space less than space 1, then the sum converges to a finite value given by the formula

S subscript infinity equals fraction numerator a over denominator 1 minus r end fraction space comma space blank open vertical bar r close vertical bar less than 1

  • a is the first term

  • r is the common ratio

  • The formula is given on the exam formula sheet

    • So you don't need to remember it

    • But you do need to know how to use it!

Examiner Tips and Tricks

  • Always check the open vertical bar r close vertical bar less than 1 condition before calculating a sum to infinity

    • Marks may depend on showing this in your working

Worked Example

The first three terms of a geometric sequence are  6 space comma space space 2 space comma space space 2 over 3.  Show that the sum to infinity of the series exists, and then find the sum to infinity.

To show the sum to infinity exists we need to know the value of the common ratio r

We can find this by dividing a term by the term immediately before it

r equals 2 over 6 equals 1 third

This satisfies open vertical bar r close vertical bar less than 1, so the series converges and the sum to infinity exists

stretchy vertical line r stretchy vertical line bold less than bold 1, so the series converges

Now use the sum to infinity formula  S subscript infinity equals fraction numerator a over denominator 1 minus r end fraction

Here  a equals 6  and  r equals 1 third

table row cell S subscript infinity end cell equals cell fraction numerator 6 over denominator 1 minus 1 third end fraction end cell row blank equals cell fraction numerator 6 over denominator open parentheses 2 over 3 close parentheses end fraction end cell row blank equals cell 6 cross times 3 over 2 end cell end table

bold italic S subscript bold infinity bold equals bold 9

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    Author
    Reviewer
    Amber

    Author: Amber

    Expertise: Maths

    Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

    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.