S_{n} = 3n^{2} – n

**Taking n = 1, we get**

S_{1} = 3(1)^{2} - (1)

⇒ S_{1} = 3 – 1

⇒ S_{1} = 2

⇒ a_{1} = 2

**Taking n = 2, we get**

S_{2} = 3(2)^{2} – 2

⇒ S_{2} = 12 – 2

⇒ S_{2} = 10

∴ a_{2} = S_{2} – S_{1} = 10 – 2 = 8

**Taking n = 3, we get**

S_{3} = 3(3)^{2} – 3

⇒ S_{3} = 27 – 3

⇒ S_{3} = 24

∴ a_{3} = 24 – 10 = 14

So, a = 1,

d = a_{2} – a_{1} = 8 - 2 = 6

Now, we have to find the 15^{th} term

a_{n} = a + (n – 1)d

a_{n} = 2 + (n – 1)6

a_{n} = 2 + 6n – 6

a_{n} = - 4 + 6n

**Hence, the n**^{th} term is 4n - 3.