A geometric progression is a sequence of numbers where each subsequent term is found by multiplying the previous term by a fixed ratio. The first term is denoted by a and the common ratio by r. The nth term is given by arn-1. The sum of the first n terms is given by (1 - rn) / (1 - r). The behavior of the sequence depends on the value of r, determining whether the terms grow, decay, or alternate in sign. Examples demonstrate calculating individual terms and sums of geometric progressions.

2. In Mathematics, a geometric
progression, also known as a
geometric sequence, is a sequence
of numbers where each term after
the first is found by multiplying the
previous one by a fixed, non-zero
number called the common ratio. It
is denoted by the letter “r”.
In a G.P, the first term of a
sequence is denoted by the letter
“a”

3. A geometric sequence is a sequence
such that any element after the first is
obtained by multiplying the preceding
element by a constant called
the common ratio which is denoted
by r. The common ratio (r) is obtained
by dividing any term by the preceding
term, i.e.,
The geometric sequence is
sometimes called the geometric
progression or GP, for short.

4. Let’s take a look at the formulae for finding the n-th
term of a G.P
If the first term is denoted by a, and the common ratio by r,
the series can be written as:
a +
e.g. 5 + 10 + 20 + 40 + …
Hence the nth term is given by:
1
 n
n aru
or 2 – 4 + 8 –16 + …
ar2+ar + ar3 + …

5. The sum of the first n terms, Sn is found as follows:
Sn = a + ar + ar2 + ar3 +…arn–2 + arn–1 …(1)
Multiply throughout by r:
r Sn = ar + ar2 + ar3 + ar4 + …arn–1 + arn …(2)
Now subtract (2) – (1): r Sn – Sn = arn – a
Factorise: Sn (r – 1) = a (rn – 1 )
Hence:
( 1)
1
n
n
a r
S
r




6. The behaviour of a geometric sequence depends on the
value of the common ratio.
If the common ratio is:
 Positive, the terms will all be the same sign as the
initial term.
 Negative, the terms will alternate between positive
and negative.
 Greater than 1, there will be exponential
growth towards positive or negative
infinity (depending on the sign of the initial term).
 1, the progression is a constant sequence.
 Between −1 and 1 but not zero, there will
be exponential decay towards zero.
 −1, the progression is an alternating sequence
 Less than −1, for the absolute values there is
exponential growth towards (unsigned) infinity, due
to the alternating sign.

7. A GP of 3 terms : a/r, a, ar
A GP of 4 terms : a/r3, a/r, ar, ar3
And so on.
If all terms in a GP are multiplied or divided
by the same number, or are raised to the
same power, then the resulting sequence is
still a GP.

8.  If a, b, c are in GP then b2=ac and b is called
the GM of a and c.
 Conversely, if b2=ac, then a, b, c are in GP.
 Sum of infinite terms of a GP:
If -1<r<1, then GP is said to converge, that is
to say that sum of infinite terms of such a GP
tends to a constant value.

9. 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 = arn–1 u10 = 2(39) = 39366
( 1)
b) Using
1
n
n
a r
S
r



8
8
2(3 1)
3 1
S



= 6560

10. Example 2: For the series 32 – 16 + 8 – 4 + 2 …
Find a) The 12th term.
b) The sum of the first 7 terms.
a) For the series, we have: a = 32, r =
Using: un = arn–1 u12 = 32
( 1)
b) Since
1
n
n
a r
S
r



71
2
7 1
2
32(1 ( ) )
1 ( )
S
 

  = 21.5
1
2
–
( )
1
2
–
11
=
1
64
–
We can write this as:
(1 )
1
n
n
a r
S
r




11. Example 3: 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 i.e. u3 = 36
The sixth term is 121.5 i.e. u6 = 121.5
Using: un = arn–1 ar2 = 36 ….(1)
ar5 = 121.5 ….(2)
Now, divide equation (2) by equation (1):
36
5.121
2
5

ar
ar
So r3 = 3.375
3
375.3r 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 = 35469
(To the nearest integer)

12. Example 4: 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:
3
4
i.e.


x
x
r
x
x
r
3
and


x
x
x
x 3
3
4 



Now, x (4x) = (x + 3)(x + 3)
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
4x2 = x2 + 6x + 9

13. ( 1)
1
n
n
a r
S
r



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:
The sum of the first n terms is given by:
1
 n
n aru
In problems where r < 1 it is better to write the above as:
(1 )
1
n
n
a r
S
r


