Let’s consider a to be the first term and d be the common difference.
And we know that, sum of first n terms is:
Sn = \(\frac{n}{2}\)(2a + (n − 1)d) and nth term is given by: an = a + (n – 1)d
Now, from the question we have
S10 = 120
⟹ 120 = \(\frac{10}{2}\)(2a + (10 − 1)d)
⟹ 120 = 5(2a + 9d)
⟹ 24 = 2a + 9d …. (1)
Also given that, a10 = 21
⟹ 21 = a + (10 – 1)d
⟹ 21 = a + 9d …. (2)
Subtracting (2) from (1), we get
24 – 21 = 2a + 9d – a – 9d
⟹a = 3
Now, on putting a = 3 in equation (2), we get
3 + 9d = 21
9d = 18
d = 2
Thus, we have the first term(a) = 3 and the common difference(d) = 2
Therefore, the nth term is given by
an = a + (n – 1)d = 3 + (n – 1)2
= 3 + 2n -2
= 2n + 1
Hence, the nth term of the A.P is (an) = 2n + 1.