Class 11 Maths Sequences Series | nth term of Arithmetic Progression |

**nth term of Arithmetic Progression**

Let a_{1}, a_{2}, a_{3}, . . . be an AP whose first term a1 is a and the common difference is d., thus the AP is a, a + d, a + 2d, a + 3d,

So, the nth term a_{n} of the AP with first term ‘a’ and common difference d is given by a_{n} = a + (n – 1) d.

a_{n} is also called the general term of the AP

Operation on an A.P. :

(i) If a constant is added to each term of an A.P., the resulting sequence is also an A.P.

(ii) If a constant is subtracted from each term of an A.P., the resulting sequence is also an A.P.

(iii) If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

(iv) If each term of an A.P. is divided by a non-zero constant then the resulting sequence is also an A.P.

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**Example 1: **Find the 10th term of the AP : 2, 7, 12, . . .

**Solution: **Here, a = 2, d = 7 – 2 = 5 and n = 10.

We have a_{n} = a + (n – 1) d

So, a_{10} = 2 + (10 – 1) × 5 = 2 + 45 = 47

Therefore, the 10th term of the given AP is 47.

**Example 2: **Determine the AP whose 3rd term is 5 and the 7th term is 9.

**Solution: **a_{3 }= a + (3 – 1) d = a + 2d = 5 --(i)

a_{7} = a + (7 – 1) d = a + 6d = 9 --(ii)

Solving the pair of linear equations (i) and (ii), we get a = 3, d = 1

Hence, the required AP is 3, 4, 5, 6, 7 . . .

**Example 3 : **Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .

**Solution: **d**= **a_{2}– a = 6 & a=5

Let 301 be a term, say, the nth term of this AP.

a_{n} = a + (n – 1) d

or 301 = 5 + (n – 1) × 6

301 = 6n – 1

Or n = 302/6 which is a fraction. Thus 301 is not in the series.

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