The series of a sequence is the sum of the sequence to a certain number of terms. It is often written as Sn. So if the sequence is 2, 4, 6, 8, 10, ... , the sum to 3 terms = S3 = 2 + 4 + 6 = 12.
This is best explained using an example:
4
S 3r
r = 1
This is 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 Ur
r = 1
This is the general case. For the sequence Ur, this means the sum of the terms obtained by substituting in 1, 2, 3, ... n in turn for r in Ur. In the above example, Ur = 3r + 2 and n = 3.
An arithmetic progression is a sequence which increases by a common difference, d, and which has a first term, a . For example: 3, 5, 7, 9, 11, is an arithmetic progression where a = 3 and 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, ... Un = 3 + 2(n - 1) = 2n + 1, which we already knew.
Sn = ½ n [ 2a + (n - 1)d ]
Example:
Sum the first 20 terms of the sequence: 1, 3, 5, 7, 9, ... (ie the first 20 odd numbers).
S20 = ½ (20) [ 2 × 1 + (20 - 1)×2 ]
= 10[ 2 + 19 × 2]
= 10[ 40 ]
= 400
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:
arn-1
For example, in the following geometric progression, the first term is 1, and the common ratio is 2:
1, 2, 4, 8, 16, ...
The sum of a geometric progression is:
a(1 - rn)
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 - 25)
1 - 2
= 2( 1 - 32)
-1
= 62
The sum to infinity of a geometric progression
In geometric progressions where |r| < 1 , the sum of the sequence as n tends to infinity approaches a 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.
