Arithmetic progressions are a series of numbers, where each term is a fixed number plus the previous term. This article discusses formulas and their uses.
An arithmetic progression is a series of numbers, each one differing from the previous number in the series by a common difference. There are two common formulas associated with arithmetic progressions. The first is the formula for the "n"th term (e.g. the sixth term) in the series, and the other is the formula to sum the series from the "n"th to the "m"th (e.g. sum the series from the third number in the series to the tenth.)
Arithmetic Progression Drummers Drumming - Grafixar
Arithmetic Progressions Examples and General Terms
The best-known of these series is the "counting numbers" i.e. 1, 2, 3, 4, ... etc. Other examples are
Example 1: 1, 2, 3, 4, ...
Example 2: 2, 5, 8, 11, ...,
Example 3: -10, 0, 10, 20 ...
The first term in the series is often called "a", and the difference between each successive number is often denoted by "d". In Example 1, the first term is 1, and the common difference, d, is also 1. In the second example, the first term, a, is 2, and the common difference is 3. In example 3, the first term is -10 and the common difference is 10.
Arithmetic Progression Formula For "n"th Term
It is easy to state what the next term is for examples 1 to 3, but what if the first term was 0.63, the common difference was 1.4, and the 8th term was needed? It is therefore useful to know the general formula for the "n" term of the progression.
The "n"th term can be derived as follows:
- the first term is "a"
- the "n"th term is the first term plus (n-1) common differences. e.g. In example 1, the third term is two numbers higher than the first, and so is larger by 2d.
- the formula for the "n"th term is therefore a + (n-1).d
To use the example with a = 0.63 and d = 1.4, and find the 8th term:
20th term = 0.63 + (8 - 1) × 1.4
= 0.63 + 7 × 1.4
= 0.63 + 9.8
= 10.61
Arithmetic Progression Formula For Sum To The "n"th Term
It is easy to add the first three terms of example 1: 1 + 2 + 3 = 6. Is there an easy way to add all the numbers from 1 to 12? (e.g. To find out how many gifts "my true love gave to me " on the 12th day of Christmas). The formula is easy to derive, and will be explained first using an example.
Using the "12 Days of Christmas" example, there are 12 days in total. On the twelfth day, the recipient gets 1 partridge, 2 turtle doves, 3 French hens, ... 12 drummers drumming. The sum of the gifts on the 1st and 12th day is 13. The sum of the gifts on the 2nd and 11th days is 13, as is the sum for the 3rd and 10th days. There are 12 days, so there are 6 "day pairs", each adding up to 13. So the number of gifts is (6 pairs of days) × 13 gifts for each pair of days = 78 gifts.
Arithmetic Progression Formula Derivation For Sum To The "n"th Term
Having used one specific example, the general proof is easier to follow. The first term is a, and the "n"th term is a + (n-1).d and added together, these give
a + (a + (n-1).d)
= 2a + (n-1).d
There are n/2 of these pairs, so the total equals
n/2 × [2a + (n-1).d]
= n.[a + (n-1).d/2]
For the days of Christmas rhyme, a = 1, d = 1, and n = 12, so the total is
12(1 +11×1 / 2)
= 12 + 66
= 78
Arithmetic Progression Formula Summary
Arithmetic progressions are number series where each number differs from the number before it by a common difference. The derivations of the two main formulas have been described, as well as some examples of using the formulas. Arithmetic progressions are often studied and used along with geometric progressions.
Arithmetic Progression Formula References
Each formula has been derived from first principles, although the article Sequences of numbers in generalized arithmetic and geometric progressions. describes both arithmetic and geometric progressions in greater detail.
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