If sn, the sum of first n terms of an AP is given by sn= (3n2-4n), then find its nth term.
Let the sum of first n terms of the A.P.=Sn
Given: Sn = 3n2 – 4n ...(i)
Now,
Replacing n by (n –1) in (i), we get,
Sn – 1 = 3(n – 1)2 – 4(n – 1)
nth term of the A.P. an = Sn – Sn – 1
∴ an = (3n2 – 4n) – [3(n – 1)2 – 4(n – 1)]
⇒ an = 3 [ n2 – (n – 1)2] – 4 [n – (n – 1)]
⇒ an = 3 (n2 – n2 + 2n – 1) – 4 (n – n + 1)
⇒ an = 3(2n –1) – 4
⇒ an = 6n – 3 – 4
⇒ an= 6n – 7
Thus, the nth term of the A.P = 6n – 7.