Question

The sum of the first n terms of an AP is given by Sn, = 3n^2 - 4n. Determine the AP and the 12th term.

Answer 1

$Sn = 3 {n}^{2} - 4n \\ first \: \: term \: t1 = S1 = 3 \times {1}^{2} - 4 \times 1 = - 1 \\ S2 = 3 \times {2}^{2} - 4 \times 2 = 4 \\ t2 = S2 - t1 = 4 + 1 = 5 \\ common \: \: difference \: \: d = t2 - t1 = 5 + 1 = 6 \\ \\ therefore \: \: ap \: \: is \\ - 1,5,11,17,..... \\ \\ t12 = t1 + (12 - 1) \times d \\ = - 1 + 11 \times 6 = 65$