11 SL notes 2015-16
Unit 1: Algebra
Lesson 5: Geometric Series
How to generate the terms of a sequence using your calculator:
https://www.youtube.com/watch?v=EK2lzsVMsG4
A geometric series is the sum of the terms of a geometric sequence.
The sum of the first n terms of a geometric series is given by 𝑆𝑛 .
For a geometric series:
𝑆𝑛 = 𝑢1 + 𝑢1 𝑟 + 𝑢1 𝑟 2 + 𝑢1 𝑟 3 + ⋯
(1)
Multiplying every term by r gives:
𝑟𝑆𝑛 = 𝑢1 𝑟 + 𝑢1 𝑟 2 + 𝑢1 𝑟 3 + 𝑢1 𝑟 4 + ⋯ (2)
(1)-(2)= 𝑆𝑛 − 𝑟𝑆𝑛 = 𝑢1 − 𝑢1 𝑟 𝑛
Subtracting
⟹ 𝑆𝑛 (1 − 𝑟) = 𝑢1 (1 − 𝑟 𝑛 )
⟹ 𝑆𝑛 =
𝑢1 (1−𝑟 𝑛 )
(1−𝑟)
This formula works for all values of 𝑟 except 𝑟 = 1.
It is most useful when 𝑟 < 1, because most terms in the sequence will be
positive.
𝑆𝑛 =
𝑢1 (1−𝑟 𝑛 )
(1−𝑟)
is the sum of a geometric series. Use when 𝑟 < 1.
The same formula will work when 𝑟 > 1, but there is another equivalent formula
which is easier to use in this situation:
Multiplying the top and bottom of the formula 𝑆𝑛 =
𝑆𝑛 =
𝑢1
(𝑟 𝑛 −1)
(𝑟−1)
𝑢1 (1−𝑟 𝑛 )
(1−𝑟)
by −1 gives
is the sum of a geometric series. Use when 𝑟 > 1.
Both formulae will work for any value of 𝑟 (except 𝑟 = 1), but it makes the
arithmetic easier if you use the correct version.
Example:
Find the sum of the first 9 terms of the sequence 5, 10, 20, 40, …
Solution:
𝑢1 = 5 and 𝑟 = 2
𝑢1 (1 − 𝑟 𝑛 )
𝑆𝑛 =
(1 − 𝑟)
5(29 − 1)
𝑆9 =
(2 − 1)
= 2555
Example: For the geometric sequence with 𝑢1 = 24 and 𝑟 = −0.5, find the sum of
the first 10 terms, giving your answer to 4 d.p.
Solution: 𝑆𝑛 =
𝑢1 (1−𝑟 𝑛 )
(1−𝑟)
𝑆10 =
24(1 − (−0.5)10 )
(1 − (−0.5))
24(1−0.510 )
=
1.5
= 15.984375
= 15.9844 to 4 d.p.
11 SL notes 2015-16
Example: the 5th term of a GP is 45, the 7th term is 135, and the sum of the first
ten terms is positive.
Find the first term and the common ratio.
1210
Show that the sum of the first ten terms is
.
√3−1
Solution:
𝑢7 = 𝑢1 𝑟 6 = 135
𝑢5 = 𝑢1 𝑟 4 = 45
𝑢
Dividing gives𝑢7 = 𝑟 2
5
𝑟2 = 3
𝑟 = ±√3
Also, since 𝑟 2 = 3, and 𝑢5 = 45, then 𝑢1 = 5.
10
If 𝑟 = √3, then
𝑆10 =
=
=
5(√3
−1)
(√3−1)
5(243−1)
√3−1
1210
√3 − 1
Which is the required answer. We must still check the other possibility.
If 𝑟 = −√3, then
𝑆𝑛 =
=
5(1−(−(√3))10 )
(1−(−√3))
5(1−243)
1+√3
which is negative, so not a possible
solution given that the sum must be positive.
So 𝑢1 = 5, 𝑟 = √3 and 𝑆10 =
1210
.
√3−1
More difficult example:
The sum of the first n terms of a series is 3𝑛 − 1. Show that the terms of
this series are in geometric progression and find the first term, and the
sum of the second set of n terms of this series. What is the sum of the
second set of 6 terms?
𝑢𝑛 = 𝑆𝑛 − 𝑆𝑛−1 This is a useful trick!
⟹ 𝑢𝑛 = (3𝑛 − 1) − (3𝑛−1 − 1)
= 3𝑛 − 3𝑛−1
= 3𝑛−1 (3 − 1)
= (3𝑛−1 )2
Hence 𝑢𝑛+1 = (3𝑛 )2
Therefore
𝑢𝑛+1
𝑢𝑛
(3𝑛 )2
= (3𝑛−1 )2
=3
Since this is a constant, it means there is a common ratio between consecutive
terms, so this is a geometric progression.
𝑢1 = (31−1 )2
=2
11 SL notes 2015-16
Consider 𝑢1 + 𝑢2 + 𝑢3 + ⋯ + 𝑢𝑛 + 𝑢𝑛+1 + 𝑢𝑛+2 + ⋯ + 𝑢2𝑛
This is the sum of the second set of n terms.
Sum of the second set of n terms = 𝑆2𝑛 − 𝑆𝑛 another useful trick!
= (32𝑛 − 1) − (3𝑛 − 1)
= 32𝑛 − 3𝑛
Hence, the sum of the second set of six terms is 312 − 36 = 530712.
Now try Exercise 6G1 on page 171.
Compound interest is an application of a geometric sequence.
Example:
Find the value after four years if $5000 is invested at 7%p.a. interest,
compounded annually.
Solution:
The initial investment of $5000 is 𝑢1 , the ratio is 1.07, because the investment
increases by 7% per annum. Since the time starts at t=0, the time of the initial
investment, then the formula for the value of the investment after t=4 years is
Value = 5000(1.07)4 = $6553.98
Example:
Find the value if $2000 is invested at 6% p.a. for 5 years, compounded monthly.
Solution:
6
Since the compounding is monthly the %age increase per month is 12, and the
number of compounding periods is 6 × 12.
Value = 2000(1.005)60 = $2697.70
Money is written to 2DP unless otherwise indicated.
Exercise 6D.3 on page 165.