Series
The series of a sequence is the sum of the sequence to a certain number of terms. It is often written as S_{n}. So if the sequence is 2, 4, 6, 8, 10, ... , the sum to 3 terms = S_{3 }= 2 + 4 + 6 = 12.
The Sigma Notation
The Greek capital sigma, written S, is usually used to represent the sum of a sequence. This is best explained using an example:
This means replace the r in the expression by 1 and write down what you get. Then replace r by 2 and write down what you get. Keep doing this until you get to 4, since this is the number above the S. Now add up all of the term that you have written down.
This sum is therefore equal to 3×1 + 3×2 + 3×3 + 3×4 = 3 + 6 + 9 + 12 = 30.
3
S 3r + 2
r = 1
This is equal to:
(3×1 + 2) + (3×2 + 2) + (3×3 + 2) = 24 .
The General Case
n
S U_{r}
r = 1
This is the general case. For the sequence U_{r}, this means the sum of the terms obtained by substituting in 1, 2, 3,... up to and including n in turn for r in U_{r}. In the above example, U_{r} = 3r + 2 and n = 3.
Arithmetic Progressions
An arithmetic progression is a sequence where each term is a certain number larger than the previous term. The terms in the sequence are said to increase by a common difference, d.
For example: 3, 5, 7, 9, 11, is an arithmetic progression where d = 2. The nth term of this sequence is 2n + 1 .
In general, the nth term of an arithmetic progression, with first term a and common difference d, is: a + (n  1)d . So for the sequence 3, 5, 7, 9, ... U_{n} = 3 + 2(n  1) = 2n + 1, which we already knew.
The sum to n terms of an arithmetic progression
This is given by:

S_{n} = ½ n [ 2a + (n  1)d ]
You may need to be able to prove this formula. It is derived as follows:
The sum to n terms is given by:
S_{n} = a + (a + d) + (a + 2d) + … + (a + (n – 1)d) (1)
If we write this out backwards, we get:
S_{n} = (a + (n – 1)d) + (a + (n – 2)d) + … + a (2)
Now let’s add (1) and (2):
2S_{n} = [2a + (n – 1)d] + [2a + (n – 1)d] + … + [2a + (n – 1)d]
So S_{n} = ½ n [2a + (n – 1)d]
Example
Sum the first 20 terms of the sequence: 1, 3, 5, 7, 9, ... (i.e. the first 20 odd numbers).
S20 = ½ (20) [ 2 × 1 + (20  1)×2 ]
= 10[ 2 + 19 × 2]
= 10[ 40 ]
= 400
Geometric Progressions
A geometric progression is a sequence where each term is r times larger than the previous term. r is known as the common ratio of the sequence. The nth term of a geometric progression, where a is the first term and r is the common ratio, is:
 ar^{n1}
For example, in the following geometric progression, the first term is 1, and the common ratio is 2:
1, 2, 4, 8, 16, ...
The nth term is therefore 2^{n1}
The sum of a geometric progression
The sum of the first n terms of a geometric progression is:

a(1  r^{n })
1 – r
We can prove this as follows:
S_{n} = a + ar + ar^{2} + … + ar^{n1} (1)
Multiplying by r:
rS_{n} = ar + ar^{2} + … + ar^{n }(2)
(1) – (2) gives us:
S_{n}(1 – r) = a – ar^{n} (since all the other terms cancel)
And so we get the formula above if we divide through by 1 – r .
Example
What is the sum of the first 5 terms of the following geometric progression: 2, 4, 8, 16, 32 ?
S5 = 2( 1  2^{5})
1  2
= 2( 1  32)
1
= 62
The sum to infinity of a geometric progression
In geometric progressions where r < 1 (in other words where r is less than 1 and greater than –1), the sum of the sequence as n tends to infinity approaches a value. In other words, if you keep adding together the terms of the sequence forever, you will get a finite value. This value is equal to:

a
1 – r
Example
Find the sum to infinity of the following sequence:
1  ,  1  ,  1  , 
1 
, 
1 
, 
1 
,  ... 
2  4  8  16  32  64 
Here, a = 1/2 and r = 1/2
Therefore, the sum to infinity is 0.5/0.5 = 1 .
So every time you add another term to the above sequence, the result gets closer and closer to 1.
Harder Example
The first, second and fifth terms of an arithmetic progression are the first three terms of a geometric progression. The third term of the arithmetic progression is 5. Find the 2 possible values for the fourth term of the geometric progression.
The first term of the arithmetic progression is: a
The second term is: a + d
The fifth term is: a + 4d
So the first three terms of the geometric progression are a, a + d and a + 4d .
In a geometric progression, there is a common ratio. So the ratio of the second term to the first term is equal to the ratio of the third term to the second term. So:
a + d = a + 4d
a a + d
(a + d)(a + d) = a(a + 4d)
a² + 2ad + d² = a² + 4ad
d²  2ad = 0
d(d  2a) = 0
therefore d = 0 or d = 2a
The common ratio of the geometric progression, r, is equal to (a + d)/a
Therefore, if d = 0, r = 1
If d = 2a, r = 3a/a = 3
So the common ratio of the geometric progression is either 1 or 3 .
We are told that the third term of the arithmetic progression is 5. So a + 2d = 5 . Therefore, when d = 0, a = 5 and when d = 2a, a = 1 .
So the first term of the arithmetic progression (which is equal to the first term of the geometric progression) is either 5 or 1.
Therefore, when d = 0, a = 5 and r = 1. In this case, the geometric progression is 5, 5, 5, 5, .... and so the fourth term is 5.When d = 2a, r = 3 and a = 1, so the geometric progression is 1, 3, 9, 27, ... and so the fourth term is 27.