The sum of first n terms of an AP is 3n²−n. Find the 25th term of this AP.

Given Sn = 3n²−n
Then Sn−1 = 3(n−1)² − (n−1)
an = Sn − Sn−1 ⇒ an = [3n²−n] − [3(n−1)² − (n−1)] = 3[n² − (n−1)²] − n + n−1 = 3(n² − n² + 2n − 1) − 1 = 3(2n − 1) − 1 = 6n − 4
Then, a25 = 6 × 25 − 4 = 150 − 4 = 146