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 * *_{ n } – S * *_{ 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.**