Arithmetic Series
Consider the following example:
A military unit purchases 10 spare parts during the first month of a contract, 15 spare
parts in the second month,
20 spare parts in the third month, 25 spare parts in the
fourth month, and so on.
The acquisition officer wants to know the total number of spare
parts the unit will
have acquired after 50 months.
This sequence of number of parts purchased in each month is called an
Arithmetic Series and the sum of this series (i.e., the total number of
purchased spare parts) can be
written as follows.
S = 10 + [10+(1*5)] + [10+(2*5)] + [10+(3*5)] + ... +[10+(49*5)]
In an Arithmetic Series, there is a fixed difference
between successive terms. In the example above, the difference between successive terms is 5 and the initial term is 10 and the 50th term is 10+(50-1)5 = 255. If the number of terms is very large, then it is difficult to compute the
above sum without using a formula. Fortunately, there is a simple formula for finding the sum of an Arithmetic Series.
In general, the sum of an Arithmetic Series can be written as follows:
S = i + (i+d) + (i+2d) + (i+3d) + ... +[i+(n-1)d]
and can be simplified as:
S = (n / 2) * [2i + (n-1) * d]
where-
i is the initial term,
d is the difference between successive terms and
n is the number of terms in the series.
Applying this sum formula to the above example, we get
S = (50 / 2) [2 * 10 + (50-1) * 5] = 6625.
Example:
Let's consider the following series:
[1, 2, 3, 4, ..., n]
This is an Arithmetic Series because the difference between successive terms is constant (1 in this case). The initial term is 1. By applying the above formula, we get the sum as follows:
S = [n * (n+1) / 2]
where n = number of terms.
We can now find the above sum for different values of n.
-
For n = 1, S = 1(1+1)/2 = 1
For n = 2, S = 2(2+1)/2 = 3
For n = 3, S = 3(3+1)/2 = 6