Geometric Sequences & Series
This week the focus is on finding the sum of a
geometric series using a formula.
We will show how to prove the formula and also
how to apply it to a range of problems.
Geometric Sequences & Series
CONTENTS:
Sum of a Geometric Series
Proof of Sum Formula
Example 1
Example 2
Example 3
Example 4
Assignment
Geometric Sequences & Series
Sum of a Geometric Series
To find the sum of a geometric series we use a
formula:
a(rn – 1)
Sn = --------- (if r > 1)
r - 1
or
a(1 - rn)
Sn = --------1 - r
(if r < 1)
Geometric Sequences & Series
Sum of a Geometric Series – Proof
Sn = a + ar + ar2 + ar3 + ... + arn-1
(1)
rSn = ar + ar2 + ar3 + ar4 + ... + arn
(2)
Subtracting (1) and (2) gives:
Sn – rSn = a – arn
all but the first and last terms cancel out
Sn(1 – r) = a(1 - rn)
factorise out Sn
Sn = a(1 – rn) / (1 – r)
bring the (1 – r) to the other side
Geometric Sequences & Series
Example 1:
Find the sum of the geometric series:
1 + 2 + 4 + 8 + ... (8 terms)
Solution:
To use the Sum formula we need to work out a, r and n.
a = 1
r = 2/1 = 2
n = 8
As r > 1 we will use the formula Sn = a(rn – 1)/ (r – 1)
Substituting into the formula gives:
S8 = 1(28 – 1)/ (2 – 1)
S8 = 1(256 – 1)/ (1)
S8 = 255
Geometric Sequences & Series
Example 2:
r
10
1
Work out  6   
2
r 1
Solution:
Remember Σ means sum. So we are trying to find the sum of
a series.
Work out the first few terms of the series but substituting in
r = 1, r = 2, r = 3 to get:
6 x (½)1 + 6 x (½)2 + 6 x (½)3 + ... = 6[(½) + (½)2 + (½)3 + ...]
(½) + (½)2 + (½)3 + ... is a geometric series with a = ½ r = ½
We have 10 terms as we want the sum from r = 1 to r = 10.
Therefore n = 10
continued on next slide
Geometric Sequences & Series
We then have a = ½ , r = ½ and n = 10
a(1 r n )
We substitute into the sum formula Sn 
because r <
1 r
1 and we will need to multiply our answer by 6.
 1   1 10  
  1     
 2   2  
Sn  6 

1
1


2




1
1 


 1 
2  1024  

6


1


2


 1  1023  

 
2  1024  
 1023

6
 6
 5.994



1
1024


2


Geometric Sequences & Series
Example 3:
Jane invests £4000 at the start of every year. She
negotiates a rate of interest of 4% per annum, which is paid
at the end of the year. How much is her investment worth at
the end of the 10th year?
Solution:
Let a be the initial investment. This gives a = 4000. The
rate of interest is 4% p.a therefore r = 1.04.
End of Year 1:
Start of Year 2:
End of Year 2:
4000 x 1.04
4000 x 1.04 + 4000
(4000 x 1.04 + 4000) x 1.04
= 4000 x 1.042 + 4000 x 1.04
= 4000(1.042 + 1.04)
continued on next slide
Geometric Sequences & Series
If we continue this pattern we get:
4000(1.0410 + 1.049 + 1.048 + ... + 1.04)
1.0410 + 1.049 + 1.048 + ... + 1.04 is a geometric series
with a = 1.04, r = 1.04 and n = 10.
We can find the sum of this series using our sum formula and
multiplying the answer by 4000 to get the total worth of the
investment at the end of the 10 years.
continued in next slide
Geometric Sequences & Series


1.04 1.0410  1 
S10  4000

1.04

1


1.040.4802442849
 4000

0.04

 0.4994540563
 4000
 49945.41

0.04


At the end of 10 years Jane’s investment is worth £49 945.41
Geometric Sequences & Series
Example 4:
Find the least value of n such that the sum 5 + 4.5 + 4.05 +
... to n terms would first exceed 45.
Solution:
This question is asking us to find n such that Sn > 45.
From the question we see that a = 1 and r = 4.5/5 = 0.9.
We can substitute these values into the sum formula to get
an inequality in n to solve.
continued in next slide
Geometric Sequences & Series
 51 - (0.9)n 

  45
 1 - 0.9 
Substituting the known values into the
sum formula and letting it be > 45.
 51 - (0.9)n 

  45
0.1


5(1 (0.9)n )  4.5
Multiplying across by 0.1
1- (0.9)n  0.9
Divide across by 5
- (0.9)n  - 0.1
(0.9)n  0.1
nlog(0.9)  log (0.1)
n  21.85
Take logs of both sides and bring power
of n forward
This implies that when n = 22 the sum of
the series will first exceed 45.
Geometric Sequences & Series
ASSIGNMENT
This weeks assignment is in Moodle. There are 5
questions to answer.
Deadline for submitting assignments is 5:00pm on
Monday 29th March.