Geometric progression
In Mathematics, a sequence is a set of ordered real numbers. We usually
write: a1 , a2 , a3 ,..., an ,...
Each one of these numbers is called term (also called element or member).
Examples:
a) 1, 3, 5, 7, 9, … is the sequence of odd numbers.
b) 7, 14, 21, 28, 35, …is the sequence of multiples of 7.
c) 1, 1, 2, 3, 5, 8, 13, … is a very famous one: the Fibonacci sequence. Here each
number is the addition of the two preceding terms.
d) 1, 4, 9, 16, 25, 36, 49, … is the sequence of natural numbers’ squares.
The nth term of a sequence is an algebraic expression which allows us to calculate
each term if we know its place in the sequence. It’s written a n .
In the previous examples we have:
a) a n  2n  1
b) a n  7n
c) a n  a n 1  a n2
d) a n  n 2
A geometric progression is a sequence which has a constant ratio between terms. It
means
a 2 a3 a 4


 ... This ratio is written r .
a1 a2 a3
Examples:
e) 1, 2, 4, 8, 16, 32, … is a geometric progression. It has a constant ratio between
terms (2, in this case), so r  2 .
f) 1, -3, 9, -27, 81, … is a geometric progression with r  3 .
1 1 1
1
g) 4,2,1, , , ,... is a geometric progression with r  .
2 4 8
2
In a geometric progression, each one of the terms can be obtained multiplying the
ratio by the preceding term, so:
a2  a1 ·r
a3  a 2 ·r  (a1 ·r )·r  a1 ·r 2
a 4  a3 ·r  (a1 ·r 2 )·r  a1 ·r 3
a5  a 4 ·r  (a1 ·r 3 )·r  a1 ·r 4
…
a n  a1 ·r n 1
This is the expression of the nth term in a geometric progression.
Example:
We write the nth term and a10 in the example e) 1, 2, 4, 8, 16, … :
a n  a1 ·r n 1 , we substitute a1  1, r  2 and calculate:
a n  1·2 n 1
a n  2 n 1
a n  a1 ·r n 1 , we substitute a1  1, r  2 , n  10 and calculate:
a10  1·2101
a10  2 9
a10  512
The sum, S n  a1  a2  a3  ...  an of the n first terms in a geometric progression,
can be calculated:
Sn 
a1 ·(r n  1)
r 1
Proof:
We write S n , S n ·r and we subtract these two expressions:
S n ·r  a2  a3  ...  an1  an  an ·r
- S n  a1  a2  a3  ...  an1  an
S n ·r  S n  a1  an ·r
S n ·(r  1)  an ·r  a1 , so:
Sn 
an ·r  a1
, using the expression of the nth term, we have:
r 1
Sn 
(a1 ·r n 1 )·r  a1
a ·(r n  1)
 1
r 1
r 1
Example:
We want to know S10 in the same example (e)):
Sn 
a1 ·(r n  1)
, we substitute a1  1, n  10, r  2 and calculate:
r 1
S10 
1·(210  1)
 1023
2 1
In a geometric progression we have:
a1 ·an  a2 ·an1  a3 ·an2  ...
(*)
Because:
a1 ·an  a1 ·a1 ·r n1  a1 ·r n1
2
a 2 ·a n 1  a1 ·r ·a1 ·r n  2  a1 ·r n 1
2
a3 ·a n  2  a1 ·r 2 ·a1 ·r n 3  a1 ·r n 1
2
…
The product, Pn  a1 ·a2 ·a3 ·...·an of the n first terms in a geometric progression, can
be calculated:
Pn  (a1 ·an ) n
Proof: Pn is written in two different ways and multiplied by the second expression:
Pn  a1 ·a2 ·a3 ·...·an1 ·an
· Pn  an ·an1 ·an2 ·...·a2 ·a1
Pn  (a1 ·an )·(a2 ·an1 )·...·(an1 ·a2 )·(an ·a1 )
2
Using (*), all the n parenthesis are equal to a1 ·a n , so:
Pn  (a1 ·an ) n
2
Pn  (a1 ·an ) n
and
Example:
We want to know P10 in the same example (e)):
Pn  (a1 ·an ) n , we substitute a1  1, n  10, a10  512
P10  (1·512)10  512 5