If the 10th term of an A.P. is 21 and the sum of its first ten terms is 120, find its nth term.

a10 = a + 9d

a + 9d = 21 (i)


S10 = [2a + 9d]


120 = 5 [2a + 9d]


24 = 2a + 9d


24 = 2 (21 – 9d) + 9d [From (i)]


-18 = -9d


d = 2


Putting the value of d in (i), we get


a = 21 – 9(2)


= 3


The nth term, an = a + (n – 1) d


= 3 + (n – 1) 2


= 3n + 2n – 2


= 2n + 1


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