Question

The sum of the first n terms of an AP is given by Sn = ( 3n^2 - n ). Find its

i ) nth term,

ii ) first term and

iii) common difference

Answer 1

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.


I hope it is helped you

Answer 2

Ignore the cancelled part .....
Hope it helps
:)