Question

If the sum of the first n terms of an A.P. is given by S_n = 3n² + 2n, find the nth term of the A.P.

Answer 1

a_n = S_n = S_{n − 1}

S_n = 3n² + 2n

S_{n − 1} = 3(n − 1)² + 2(n − 1)

S_{n − 1} = 3(n² − 2n + 1) + 2n − 2

S_{n − 1} = 3n² − 6n + 3 + 2n − 2

S_{n − 1} = 3n² − 4n + 1

Now, subtract S_{n − 1} from S_n

a_n = (3n² + 2n) − (3n² − 4n + 1)

a_n = 3n² + 2n − 3n² + 4n − 1

a_n = 6n − 1