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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