Sum of terms

S_n= n² + 2n

and S, (n – 1)² + 2(n-1)

We know that nth term of A.P.

T_n= S_n – S_{n – 1}

⇒ T_n= (n² + 2n) – [(n – 1)² + 2(n − 1)]

= [n² + 2n] – [n² + 1 – 2n + 2n – 2]

= [n² + 2n] – [n² – 1]

= n² + 2n – n² + 1

= 2n + 1

First term T₁= 2 × 1 + 1 = 2 + 1 = 3

Second term, T₂ = 2 2 + 1 = 4 + 1 = 5

Common difference d = T₂ – T₁ = 5 – 3 = 2

Hence, first term is 3 and common difference is 2.