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

Given: S * *_{ n } = 3 * n * ^{ 2 } â€“ 4 * n * ...(i)

Now,

Replacing * n * by ( * n * â€“1) in (i), we get,

S * *_{ n } _{ â€“ 1 } = 3( * n * â€“ 1) ^{ 2 } â€“ 4( * n * â€“ 1)

* n * ^{ th } term of the A.P. * a *_{ n } = S * *_{ n } â€“ S * *_{ n } _{ â€“ 1 }

âˆ´ * a *_{ n } = (3 * n * ^{ 2 } â€“ 4 * n * ) â€“ [3( * n * â€“ 1) ^{ 2 } â€“ 4( * n * â€“ 1)]

â‡’ * a *_{ n } = 3 [ * n * ^{ 2 } â€“ ( * n * â€“ 1) ^{ 2 } ] â€“ 4 [ * n * â€“ ( * n * â€“ 1)]

â‡’ * a *_{ n } = 3 ( * n * ^{ 2 } â€“ * n * ^{ 2 } + 2 * n * â€“ 1) â€“ 4 ( * n * â€“ * n * + 1)

â‡’ * a *_{ n } = 3(2 * n * â€“1) â€“ 4

â‡’ * a *_{ n } = 6 * n * â€“ 3 â€“ 4

â‡’ * a *_{ n } = 6 * n * â€“ 7

Thus, **the** ** ** * ***n** ^{ th } ** ** **term of the A.P = 6** * ***n** ** ** **â€“ 7.**