Sum of first n terms be S_n given as

S_n = 2n + 3n²

We have to find the r^th term that is a_r

Using the hint: the n^th term is given as a_n = S_n – S_n-1

⇒ a_r = S_r – S_r-1

Using S_n = 2n + 3n²

⇒ a_r = 2r + 3r² – (2(r – 1) + 3(r – 1)²)

⇒ a_r = 2r + 3r² – (2r – 2 + 3(r² – 2r + 1))

⇒ a_r = 2r + 3r² – (2r – 2 + 3r² – 6r + 3)

⇒ a_r = 6r – 1

Hence the r^th term is 6r – 1