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**

**Definition of Arithmetic Progression****General Form of an Arithmetic Progress****Arithmetic Mean****Sum of the First n Terms of an Arithmetic Progression****Sum of the Cubes of First n Natural Numbers****Sum of First n Natural Numbers****Sum of the Squares of First n Natural Numbers****Properties of Arithmetic Progression****Selection of Terms in an Arithmetic Progression****Arithmetic Progression Formulae****Problems on Arithmetic Progression****Problems on Sum of 'n' Terms of Arithmetic Progression**

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