|Class 11 Maths Sequences Series||nth term of Arithmetic Progression|
nth term of Arithmetic Progression
Let a1, a2, a3, . . . 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 an of the AP with first term ‘a’ and common difference d is given by an = a + (n – 1) d.
an 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.
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 an = a + (n – 1) d
So, a10 = 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: a3 = a + (3 – 1) d = a + 2d = 5 --(i)
a7 = 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= a2– a = 6 & a=5
Let 301 be a term, say, the nth term of this AP.
an = 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.