Trying to find the right answer to this..

S n = 3n 2 –4n ⇒ S n – 1 = 3 (n – 1) 2 –4 (n – 1) ∴ S n – S n –1 = (3n 2 – 4n) – {3(n – 1) 2 – 4 (n –1)} ⇒ a n = (3n 2 – 4n) – {3(n 2 – 2n+ 1) – 4n ± 4} = 3n 2 – 4n – {3n 2 – 6n + 3 – 4n + 4} = 3n 2 – 4n – {3n 2 – 10n + 7} = 3n 2 – 4n – 3n 2 + 10n – 7 = 6n – 7 Hence, required n th term is 6n – 7.

If S_n, the sum of first n terms of an A .P. is given by S_n = 3n² – 4n, then find its n^thterm.

S_n = 3n² –4n
⇒ S_{n – 1}= 3 (n – 1)² –4 (n – 1)
∴ S_n – S_{n –1}= (3n² – 4n) – {3(n – 1)² – 4 (n –1)}
⇒ a_n = (3n² – 4n) – {3(n² – 2n+ 1) – 4n ± 4} = 3n² – 4n – {3n² – 6n + 3 – 4n + 4}
= 3n² – 4n – {3n² – 10n + 7}
= 3n² – 4n – 3n² + 10n – 7
= 6n – 7
Hence, required n^th term is 6n – 7.

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