If s_{n}, the sum of first n terms of an AP is given by s_{n}= (3n^{2}-4n), then find its nth term.

## EXPERT ANSWER

Let the sum of first * n * terms of the A.P.=S * _{ n } *

Given: S * _{ n } * = 3

*n*

^{ 2 }– 4

*n*...(i)

Now,

Replacing * n * by ( * n * –1) in (i), we get,

S * _{ n } *

_{ – 1 }= 3(

*n*– 1)

^{ 2 }– 4(

*n*– 1)

* n * ^{ th } term of the A.P. * a _{ n } * = S

*– S*

_{ n }

_{ n }_{ – 1 }

∴ * a _{ n } * = (3

*n*

^{ 2 }– 4

*n*) – [3(

*n*– 1)

^{ 2 }– 4(

*n*– 1)]

⇒ * a _{ n } * = 3 [

*n*

^{ 2 }– (

*n*– 1)

^{ 2 }] – 4 [

*n*– (

*n*– 1)]

⇒ * a _{ n } * = 3 (

*n*

^{ 2 }–

*n*

^{ 2 }+ 2

*n*– 1) – 4 (

*n*–

*n*+ 1)

⇒ * a _{ n } * = 3(2

*n*–1) – 4

⇒ * a _{ n } * = 6

*n*– 3 – 4

⇒ * a _{ n } * = 6

*n*– 7

Thus, **the** ** ** * n *

^{ th }

**term of the A.P = 6**

**n**

**– 7.**

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