The general form of an Arithmetic Progress is {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, ..........}, where ‘a’ is known as the first term of the Arithmetic Progress and ‘d’ is known as the common difference (C.D.).
If a is the first term and d is the common difference of an Arithmetic Progress, then its nth term is a + (n  1)d.
Let a\(_{1}\), a\(_{2}\), a\(_{3}\), a\(_{4}\), ........, a\(_{n}\), .................. be the given Arithmetic Progress. Then a\(_{1}\) = first term = a
By the definition, we have
a\(_{2}\)  a\(_{1}\) = d
⇒ a\(_{2}\) = a\(_{1}\) + d
⇒ a\(_{2}\) = a + d
⇒ a\(_{2}\) = (2  1)a + d:
a\(_{3}\)  a\(_{2}\) = d
⇒ a\(_{3}\) = a\(_{2}\) + d
⇒ a\(_{3}\) = (a + d) + d
⇒ a\(_{3}\) = a + 2d
⇒ a\(_{3}\) = (3  1)a + d:
a\(_{4}\)  a\(_{3}\) = d
⇒ a\(_{4}\) = a\(_{3}\) + d
⇒ a\(_{4}\) = (a + 2d) + d
⇒ a\(_{4}\) = a + 3d
⇒ a\(_{4}\) = (4  1)a + d:
a\(_{5}\)  a\(_{4}\) = d
⇒ a\(_{5}\) = a\(_{4}\) + d
⇒ a\(_{5}\) = (a + 3d) + d
⇒ a\(_{5}\) = a + 4d
⇒ a\(_{5}\) = (5  1)a + d:
Similarly, a\(_{6}\) = (6  1)a + d:
a\(_{7}\) = (7  1)a + d:
a\(_{n}\) = a + (n  1)d.
Therefore, nth term of an Arithmetic Progress whose first term = ‘a’ and common difference = ‘d’ is a\(_{n}\) = a + (n  1)d.
nth term of an Arithmetic Progress from the end:
Let a and d be the first term and common difference of an Arithmetic Progress respectively having m terms.
Then nth term from the end is (m  n + 1)th term from the beginning.
Therefore, nth term of the end = a\(_{m  n + 1}\) = a + (m  n + 1  1)d = a + (m  n)d.
We can also find the general term of an Arithmetic Progress according to the process below.
To find the general term (or the nth term) of the Arithmetic Progress {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, ..........}.
Clearly, for the Arithmetic Progress is {a, a + d, a + 2d, a + 3d, ..........} we have,
Second term = a + d = a + (2  1)d = First term + (2  1) × Common Difference.
Third term = a + 2d = a + (3  1)d = First term + (3  1) × Common Difference.
Fourth term = a + 3d = a + (4  1)d = First term + (4  1) × Common Difference.
Fifth term = a + 4d = a + (5  1)d = First term + (5  1) × Common Difference.
Therefore, in general, we have,
nth term = First + (n  1) × Common Difference = a + (n  1) × d.
Hence, if the nth term of the Arithmetic Progress {a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, ..........} be denoted by t\(_{n}\), then t\(_{n}\) = a + (n  1) × d.
`Solved examples on general form of an Arithmetic Progress
1. Show that the sequence 3, 5, 7, 9, 11, ......... is an Arithmetic Progress. Find its 15th term and the general term.
Solution:
First term of the given sequence = 3
Second term of the given sequence = 5
Third term of the given sequence = 7
Fourth term of the given sequence = 9
Fifth term of the given sequence = 11
Now, Second term  First term = 5  3 = 2
Third term  Second term = 7  5 = 2
Fourth term  Third term = 9  7 = 2
Therefore, the given sequence is an Arithmetic Progress with the common difference 2.
We know that nth term of an Arithmetic Progress, whose first term is a and common difference is d is t\(_{n}\) = a + (n  1) × d.
Therefore, 15th term of the Arithmetic Progress = t\(_{15}\) = 3 + (15  1) × 2 = 3 + 14 × 2 = 3 + 28 = 31.
General term = nth term = a\(_{n}\) = a + (n  1)d = 3 + (n  1) × 2 = 3 + 2n  2 = 2n + 1
2. Which term of the sequence 6, 11, 16, 21, 26, ....... is 126?
Solution:
First term of the given sequence = 6
Second term of the given sequence = 11
Third term of the given sequence = 16
Fourth term of the given sequence = 21
Fifth term of the given sequence = 26
Now, Second term  First term = 11  6 = 5
Third term  Second term = 16  11 = 5
Fourth term  Third term = 21  16 = 5
Therefore, the given sequence is an Arithmetic Progress with the common difference 5.
Let 126 is the nth term of the given sequence. Then,
a\(_{n}\) = 126
⇒ a + (n  1)d = 126
⇒ 6 + (n  1) × 5 = 126
⇒ 6 + 5n  5 = 126
⇒ 5n + 1 = 126
⇒ 5n = 126  1
⇒ 5n = 125
⇒ n = 25
Hence, 25th term of the given sequence is 126.
3. Find the seventeenth term of the Arithmetic Progress {31, 25, 19, 13, ..................... }.
Solution:
The given Arithmetic Progress is {31, 25, 19, 13, ..................... }.
First term of the given sequence = 31
Second term of the given sequence = 25
Third term of the given sequence = 19
Fourth term of the given sequence = 13
Now, Second term  First term = 25  31 = 6
Third term  Second term = 19  25 = 6
Fourth term  Third term = 13  19 = 6
Therefore, common difference of the given sequence = 6.
Thus, the 17th term of the given Arithmetic Progress = a + (n 1)d = 31 + (17  1) × (6) = 31 + 16 × (6) = 31  96 = 65.
Note: Any term of an Arithmetic Progress can be obtained if its first term and common difference are given.
`● Arithmetic Progression
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