Transcript Geometric Progression

GEOMETRIC
PROGRESSIONS
A Geometric Progression (GP) or Geometric Series is
one in which each term is found by multiplying the
previous term by a fixed number (common ratio).
e.g. 5 + 10 + 20 + 40 + …
or
2 – 4 + 8 –16 + …
If the first term is denoted by a, and the common ratio by r,
the series can be written as:
a + ar + ar2 + ar3 + …
Hence the
nth
term is given by:
u n  ar
n 1
The sum of the first n terms, Sn is found as follows:
Sn = a + ar + ar2 + ar3 +…ar n–2 + ar n–1 …(1)
Multiply throughout by r:
r Sn = ar + ar2 + ar3 + ar4 + …ar n–1 + ar n …(2)
Now subtract (2) – (1):
Factorise:
r Sn – Sn = ar n – a
Sn (r – 1) = a (r n – 1 )
a ( r  1)
n
Hence:
Sn 
r 1
Example 1: For the series 2 + 6 + 18 + 54 + …
Find
a) The 10th term.
b) The sum of the first 8 terms.
a) For the series, we have:
a = 2, r = 3
Using: un = ar n–1
u10 = 2(3 9)
a ( r  1)
8
n
b) U sing S n 
r 1
2 (3  1)
= 39 366
S8 
3 1
= 6560
Example 2: For the series 32 – 16 + 8 – 4 + 2 …
a) The 12th term.
b) The sum of the first 7 terms.
Find
a) For the series, we have:
Using: un = ar n–1
a ( r  1)
1
–
a = 32, r =
2
11
1
1
–
–
(
)
=
u12 = 32
2
64
b) S ince S n 
r 1
a (1  r )
n
n
We can write this as: S n 
1 r
(This ensures that the denominator is positive).
3 2 (1  (  12 ) )
7
S7 
1  (  12 )
= 21.5
10

Example 3: Find : a)
12
r
6(2 )
b)
r 0

( 3  20 )
r
r 1
a) Firstly, we need to find the first few terms:
The series is: 6 + 12 + 24 + 48 + … We have a = 6, r = 2
The number of terms, n = 11
a ( r  1)
n
U sin g S n 
r 1
 S 11 
6(2
11
 1)
2 1
= 12 282
12
b)

( 3  20 ) = (31 – 20 ) + (32 – 20 ) + (33 – 20 ) + …
r
r 1
= (31 + 32 + 33 + …) – ( 20 + 20 + 20 + …)
A Geometric Series

3(3
12
 1)
3 1
– (20 × 12)
12 of these
= 797 160 – 240
= 796 920
Example 4: In a Geometric Series, the third term is 36, and the
sixth term is 121.5. For the series, find the common
ratio, the first term and the twentieth term.
The third term is 36
The sixth term is 121.5
i.e. u3 = 36
i.e. u6 = 121.5
ar 2 = 36
….(1)
ar 5 = 121.5 ….(2)
Using: un = ar n–1
Now, divide equation (2) by equation (1):
So r3 = 3.375  r  3 . 375
3
ar
5
ar
2

121 . 5
36
r = 1.5
Substitute this value into equation (1): a(1.5)2 = 36
a = 16
Now the 20th term, u20 = ar19 = 16 (1.5)19 = 35 469
(To the nearest integer)
Example 5: The first three terms of a geometric progression are
x, x + 3, 4x. Find the two possible values for the
common ratio. For each value find these first three
terms, and the common ratio.
The ratio of a G.P. is found by dividing a term by the previous term:
i.e.
Now,
r 
4x
x3
and
r 
x3

x
x (4x) = (x + 3)(x + 3)

4x
x3

x3
x
4x2 = x2 + 6x + 9
3x2 – 6x – 9 = 0
Divide by 3:
x2 – 2x – 3 = 0
(x – 3)(x + 1) = 0
So, either x = 3, giving the terms: 3, 6, 12 with ratio r = 2
or x = –1, giving the terms: –1, 2, –4 with ratio r = –2
Example 6: £100 is invested into an account (earning 5% compound
interest per annum), at the start of every year. Find the
amount in the account at the end of the 8th year.
The amount in the account at the start of each year earns 5% interest.
i.e. The amount is increased by 5%.
To increase an amount by 5%, multiply by 1.05
The amount in the account:
At the end of the 1st year = (100 × 1.05)
At the start of the 2nd year = 100 + (100 × 1.05)
At the end of the 2nd year = [100 + (100 × 1.05)] × 1.05
= (100 × 1.05) + (100 × 1.052 )
At the start of the 3rd year = 100 + (100 × 1.05) + (100 × 1.052 )
At the end of the 3rd year = {100 + (100 × 1.05) + (100 × 1.052 )} × 1.05
= (100 × 1.05) + (100 × 1.052 ) + (100 × 1.053)
The amount we now have:
At the end of the 3rd year = (100×1.05) + (100×1.052 ) + (100×1.053)
This is a GP
With a = 100 × 1.05 = 105
and r = 1.05
We want the sum to 8 terms, i.e. n = 8
a ( r  1)
n
U sin g S n 
r 1
105 (1.05  1)
8

S8 
1.05  1
= 1002.66
Hence, the amount in the account after 8 years is £1002.66
(To the nearest penny.)
Summary of key points:
A Geometric Series is one in which the terms are found by
multiplying each term by a fixed number (common ratio).
The nth term is given by:
u n  ar
n 1
The sum of the first n terms is given by:
a ( r  1)
n
Sn 
r 1
In problems where r < 1 it is better to write the above as:
a (1  r )
n
Sn 
1 r
This PowerPoint produced by R. Collins; © ZigZag Education 2008–2010