Q4 of 176 Page 8

If the sum of the first n terms of an A.P. is given by Sn = (3n2- n), find its

(i) first term (ii) common difference


(iii) nth term.

Sn = 3n2 – n


Taking n = 1, we get


S1 = 3(1)2 - (1)


⇒ S1 = 3 – 1


⇒ S1 = 2


⇒ a1 = 2


Taking n = 2, we get


S2 = 3(2)2 – 2


⇒ S2 = 12 – 2


⇒ S2 = 10


∴ a2 = S2 – S1 = 10 – 2 = 8


Taking n = 3, we get


S3 = 3(3)2 – 3


⇒ S3 = 27 – 3


⇒ S3 = 24


∴ a3 = 24 – 10 = 14


So, a = 1,


d = a2 – a1 = 8 - 2 = 6


Now, we have to find the 15th term


an = a + (n – 1)d


an = 2 + (n – 1)6


an = 2 + 6n – 6


an = - 4 + 6n


Hence, the nth term is 4n - 3.


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