A geometric sequence is one where to get from one term to the
next you multiply by the same number each time. This number
is called the common ratio, r.
1
Eg
2
3
4
2, 10, 50, 250 ...
x5
10=5
2
x5
x5
50=5
10
250=5
50
r=5
1
Eg
2
90, -30,
x
-30=
90
1

3
1
3
3
4
10,  3
x
1
3
1
3
r=
x 1
3
10 =  1
3
-30
Common ratio =u2
u1
1

3
...
3
1
3
10
= 1
3
‘Second term divided by
the first’
r= the number you times by to
get to the next term
a= the first term of the
sequence
Eg
1
2
3
4
n
2,
10,
50,
250 ...
? ...
a
ar
ar2
ar3
arn-1
This is the same for all geometric
sequences
r= 5
a=2
r= the number you times by to
get to the next term
a= the first term of the
sequence
What terms have you got/need
to find?
The second term of a geometric sequence is 4 and the 4th
term is 8. Find the values of a) the common ration b) first
term and c) the 10th term.
So what do we have:
1
2
3
?,
4,
?,
ar
4
8 ....
ar3
r= the number you times by to get to the
next term
The second term of a geometric sequence
is 4 and the 4th term is 8. Find the values
of a) the common ration b) first term and
c) the 10th term.
a= the first term of the sequence
What terms have you got/need to find?
So what do we have:
1
2
3
?,
4,
?,
ar
4
ar3
ar3 =8
ar 4
Using ar= 4
1) ar =4
2) ar3 =8
8 ....
r2 =2
a√2 = 4
a= 4
√2
r =√2
a= 2 √2
r =√2
r= the number you times by to get to the
next term
a= 2 √2
a= the first term of the sequence
What terms have you got/need to find?
the 10th term
nth term = arn-1
1
2
3
?,
4,
?,
4 ......
8 ....
9
ar =
10
ar9
9
(√2)
2 √2
9
10
ar = 2 √2
9
5
ar = 2 x2
9
6
ar = 2
th
10 term= 64
r= the number you times by to get to the
next term
Andy invests $A at a rate of interest 4%
per annum. After 5 years it will be worth
$10 000. How much (to the nearest
penny) will it be worth after 10 years?
a= the first term of the sequence
What terms have you got/need to find?
So what do we have:
r=1.04 ( think if it will increase in value you need that 1!)
a= £A (this gives you a hint that you will need to work out A)
Think about the terms:
1 year
1
a
2 year
2
?
4 year
3 year
3
?
4
?
5 year
5
?
6
10000
ar5=10 000
( Power is usually the same as the years but always check!)
ar10=?
( Using the same idea after 10 years will be the 11th term)
Andy invests $A at a rate of interest 4%
per annum. After 5 years it will be worth
$10 000. How much (to the nearest
penny) will it be worth after 10 years?
ax1.045=10 000
£A= £10 000
1.045
r= the number you times by to get to the
next term
a= the first term of the sequence
What terms have you got/need to find?
( Replace the r!)
£A = £8219.27
ar10=?
£10 000 x 1.0410 = £10 000 x 1.045
1.045
ar10= £12166.53
What is the first term in the geometric
progression 3,6,12,24.... To exceed 1
million?
r= the number you times by to get to the
next term
a= the first term of the sequence
What terms have you got/need to find?
r=2 ( 6 divided by 3)
a= 3
nth term = arn-1 > 1000 000
3x2n-1 > 1000 000
2n-1 > 1000 000
3
n-1> log2 (1000 000)
3
n-1> 18.35 (2dp)
n> 19.35 (2dp)
n = 20
So 20th term
A geometric series is the sum of a geometric sequence
Terms:
1
2
Sn
a + ar + ar2 + ar3 +.......+ ar
Terms:
1
2
3
3
4
4
n-1
n-2
n-1
n-1
n
+ arn-1
n
rSn
ar + ar2 + ar3 + ar4 +.......+ ar
Sn
rSn
a + ar + ar2 + ar3 +.......+ arn-2 + arn-1
ar + ar2 + ar3 + ar4 +.......+ arn-1 + arn
Sn- r Sn = a-arn
Sn(1-r)= a(1-rn)
Sn = a(1-rn)
1-r
+
arn
r 1
r 1
Sn = a(1-rn)
1-r
Sn = a(rn-1)
r-1
An investor invests £2000 on January 1st every year in an account that guarantees 4% per
annum, If the interest is calculated on the 31st of December each year, how much will be
in the account at the end of the 10th year?
So using logic break it down to understand what is
happening:
End of year 1:
2000 x 1.04
Start of year 2:
End of year 2:
2000 x 1.04 +2000
(2000 x 1.04 +2000) x 1.04
2000 x 1.042 +2000 x 1.04
2000 x 1.042 +2000 x 1.04 +2000
Start of year 3:
Start of year 3:
(2000 x 1.042 +2000 x 1.04 +2000) x 1.04
2000 x 1.043 +2000 x 1.042 +2000 x 1.04
End of year 1:
2000 x 1.04
Start of year 2:
End of year 2:
2000 x 1.04 +2000
(2000 x 1.04 +2000) x 1.04
2000 x 1.042 +2000 x 1.04
2000 x 1.042 +2000 x 1.04 +2000
Start of year 3:
Start of year 3:
Sn = a(rn-1)
r-1
(2000 x 1.042 +2000 x 1.04 +2000) x 1.04
2000 x 1.043 +2000 x 1.042 +2000 x 1.04
End of year 10:
2000 x 1.0410 +2000 x 1.049 +..............+2000 x 1.04
End of year 10:
2000 (1.0410 + 1.049 +..............+1.04)
End of year 10:
2000 x 1.04 (1.0410-1)
1.04-1
End of year 10:
= £24 972.70
Some problems could be described like this

10
The sum of
3

2
r 1
This for...
..... r=1 to 10
r
Sn = a(rn-1)
r-1
= 3x21 +3x22 +3x23 +3x24 +3x25 +3x26 +3x27 +3x28 +3x29 +3x210
=3(21 +22 +23 +24 +25 +26 +27 +28 +29 +210)
= 3 x 2 (210 -1)
2-1
S10 = 6138
Convergent means that the series tends towards a specific value as more
terms are added. This value is called the limit.
Consider this series S= 3 + 1.5 + 0.75 + 0.375 + ...
1 n
3(1  ( ) )
1 n
2
s
 6(1  ( ) )
1
2
1
2
Test for different values of n:
n  3, S3  5.25
n  5, S5  5.8125
n  10, S10  5.9994
n  20, S 20  5.999994
r= 1
2
a=3
As n gets larger S
becomes closer to 6
This series
S= 3 + 1.5 + 0.75 + 0.375 + ...
Is a convergent series.
This happens because: -1< r >1
If this were not true it would not be convergent.
Sn = a(1-rn)
1-r
But if -1< r >1,
r 0
n
S∞ = a(1-0)
1-r
S∞ = __a__
1-r
as
n 
nth term = arn-1
Sn = a(1-rn)
1-r
S∞ = __a__
1-r
r 1