An arithmetic progression is a sequence of numbers in which the consecutive terms (beginning with the second term) are formed by adding a constant quantity with the preceding term.
Definition of Arithmetic Progression: A sequence of numbers is known as an arithmetic progression (A.P.) if the difference of the term and the preceding term is always same or constant.
The constant quantity stated in the above definition is called the common difference of the progression. The constant difference, generally denoted by d is called common difference.
a\(_{n + 1}\)  a\(_{n}\) = constant (=d) for all n∈ N
From the definition, it is clear that an arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant.
Examples on Arithmetic Progression:
1. 2, 1, 4, 7, 10 ……………. is an A.P. whose first term is 2 and common difference is 1  (2) = 1 + 2 = 3.
2. The sequence {3, 7, 11, 15, 19, 23, 27, …………………} is an Arithmetic Progression whose common difference is 4, since
Second term (7) = First term (3) + 4
Third term (11) = Second term (7) + 4
Fourth term (15) = Third term (11) + 4
Fifth term (19) = Fourth term (15) + 4 etc.
3. The sequence {58, 43, 28, 13, 2, 17, 32, …………………} is an Arithmetic Progression whose common difference is 15, since
Second term (43) = First term (58) + (15)
Third term (28) = Second term (43) + (15)
Fourth term (13) = Third term (28) + (15)
Fifth term (2) = Fourth term (13) + (15) etc.
4. The sequence {11, 23, 35, 47, 59, 71, 83, …………………} is an Arithmetic Progression whose common difference is 4, since
Second term (23) = First term (11) + 12
Third term (35) = Second term (23) + 12
Fourth term (47) = Third term (35) + 12
Fifth term (59) = Fourth term (47) + 12 etc.
Algorithm to determine whether a sequence is an Arithmetic Progression or not when its nth term is given:
Step I: Obtain a\(_{n}\)
Step II: Replace n by n + 1 in a\(_{n}\) to get a\(_{n + 1}\).
Step III: calculate a\(_{n + 1}\)  a\(_{n}\).
When a\(_{n + 1}\) is independent of n then, the given sequence is an Arithmetic Progression. And, when a\(_{n + 1}\) is not independent of n then, the given sequence is not an Arithmetic Progression.
`The following examples illustrate the above concept:
1. Show that the sequence < a\(_{n}\)> defined by a\(_{n}\) = 2n + 3 is an Arithmetic Progression. Also fine the common difference.
Solution:
The given sequence a\(_{n}\) = 2n + 3
Replacing n by (n + 1), we get
a\(_{n + 1}\) = 2(n + 1) + 3
a\(_{n + 1}\) = 2n + 2 + 3
a\(_{n + 1}\) = 2n + 5
Now, a\(_{n + 1}\)  a\(_{n}\) = (2n + 5)  (2n + 3) = 2n + 5  2n  3 = 2
Hence, a\(_{n + 1}\)  a\(_{n}\) is independent of n, which is equal to 2.
Therefore, the given sequence a\(_{n}\) = 2n + 3 is an Arithmetic Progression with common difference 2.
2. Show that the sequence < a\(_{n}\)> defined by a\(_{n}\) = 3n\(^{2}\) + 2 is not an Arithmetic Progression.
Solution:
The given sequence a\(_{n}\) = 3n\(^{2}\) + 2
Replacing n by (n + 1), we get
a\(_{n + 1}\) = 3(n + 1)\(^{2}\) + 2
a\(_{n + 1}\) = 3(n\(^{2}\) + 2n + 1) + 2
a\(_{n + 1}\) = 3n\(^{2}\) + 6n + 3 + 2
a\(_{n + 1}\) = 3n\(^{2}\) + 6n + 5
Now, a\(_{n + 1}\)  a\(_{n}\) = (3n\(^{2}\) + 6n + 5)  (3n\(^{2}\) + 2) = 3n\(^{2}\) + 6n + 5  3n\(^{2}\)  2 = 6n + 3
Therefore, a\(_{n + 1}\)  a\(_{n}\) is not independent of n.
Hence a\(_{n + 1}\)  a\(_{n}\) is not constant.
Thus, the given sequence a\(_{n}\) = 3n\(^{2}\) + 2 is not an Arithmetic Progression.
Note: To obtain the common difference of a given arithmetic progression we required to subtract its any term from that which follow it. That is,
Common Difference = Any term  Its preceding term.
`● Arithmetic Progression
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