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