Question
If Sn, the sum of first n terms of an A .P. is given by Sn = 3n2 – 4n, then find its nth term.
Solution
Sn = 3n2 –4n
⇒ Sn – 1= 3 (n – 1)2 –4 (n – 1)
∴ Sn – Sn –1 = (3n2 – 4n) – {3(n – 1)2 – 4 (n –1)}
⇒ an = (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 nth term is 6n – 7.
