11 SL notes 2015-16
Unit 1: Algebra
Lesson 4: Geometric Sequences
A Geometric Sequence, also called a Geometric Progression or GP for short, is a
sequence of numbers where the ratio between consecutive terms is constant,
which is called the common ratio. This means that to get from one term to the
next we multiply by the common ration, 𝑟.
e.g. 4, 12, 36, 108, 324, … is a geometric sequence with first term, 𝑢1 , is 4, and
common ratio, 𝑟, equal to 3.
The general form of a geometric sequence is:
𝑢1, 𝑢1 𝑟, 𝑢1 𝑟 2 , 𝑢1 𝑟 3 , 𝑢1 𝑟 4 , …
The nth term of a geometric sequence is:
𝑢𝑛 = 𝑢1 𝑟 𝑛−1
Example 1: Find the tenth term of the geometric sequence which starts
3, 21, 147, 1029,…
Solution
21
147
= 7, and 21 = 7, so this is a geometric sequence with 𝑢1 = 3, and 𝑟 = 7.
3
To find the tenth term:
𝑢𝑛 = 𝑢1 𝑟 𝑛−1
𝑢10 = 3(7)(10−1) = 121 060 821
Question
Find the ninth term of the GP with first term 55750 and common ratio 0.2. Give
your answer correct to three significant figures.
Solution
To find the ninth term:
𝑢𝑛 = 𝑢1 𝑟 𝑛−1
𝑢𝑛 = 55750(0.2)(9−1) = 0.14272
𝑢9 = 0.143 3SF
Question
Given a geometric sequence with 𝑢12 = 885735 and 𝑟 = 3, find 𝑢1 .
Solution
To find the first term:
𝑢𝑛 = 𝑢1 𝑟 𝑛−1
𝑢12 = 𝑢1 𝑟 (12−1)
885735 = 𝑢1 311
885735
𝑢1 = 311
𝑢1 = 5
11 SL notes 2015-16
Example 2
A geometric sequence has first term 2 and fourth term 686. Find the common
ratio, 𝑟, and the eighth term.
Solution
𝑢4 𝑢1 (𝑟)(4−1)
=
= 𝑟3
𝑢1
𝑢1
𝑟3 =
686
= 343
2
3
𝑟 = √343 = 7
𝑢8 = 2(7)7 = 1647086
NB: you would NOT round this to 3SF unless told to, because it is an exact
answer.
Example 3
A sequence has a position-to-term rule 𝑢𝑛 = 7(2)𝑛 .
Prove that the sequence is a geometric progression.
Find 𝑢1 , 𝑢2 𝑎𝑛𝑑 𝑢3 .
Solution
To prove that it is a geometric sequence it is necessary to prove that there is a
common ration between consecutive terms.
𝑢𝑛 = 7(2)𝑛 and 𝑢𝑛+1 = 7(2)(𝑛+1)
𝑢𝑛+1 7(2)(𝑛+1)
=
𝑢𝑛
7(2)𝑛
=2
Hence there is a constant (common) ration between any two consecutive terms.
Therefore this is a geometric progression.
𝑢1 = 7(2)(1−1) = 7
𝑢2 = 7(2)(2−1) = 14
𝑢3 = 7(2)(3−1) = 28
Questions:
1.
In a geometric sequence the third term equals 12 and the fifth term
equals 300. Find the possible values of the common ratio.
Answer: 𝑟 = ±5
2.
A GP has 𝑢1 = 1000 and 𝑟 = 0.7.
How many terms are needed before the size of the terms goes below 1?
Answer: 21 terms
3.
Prove that the sequence whose general term is 𝑢𝑛 = 2(3)𝑛 is a GP.
Answer: see example 3 above
4.
A GP has 𝑢6 = 324 and 𝑢8 = 36. Find two the two possible values of 𝑢5 .
Answer 𝑢5 = ±4
Further Questions: check your understanding of the basics by doing a selection
of questions from Ex 6D.1 Page 160, then do the application problems about
growth and decay in Ex 6D.2 Page 163.
11 SL notes 2015-16
How to generate a sequence on your TI-84:
https://www.youtube.com/watch?v=EK2lzsVMsG4