The sum of the first n terms of an AP is given by S_n = (3n² - n) . Find its

(i) n^th term,

(ii) first term and

(iii) Common difference.

(i) Let a_n be the n^th term of the AP.

To find: a_n

Then a_n =S_n - S_{n - 1}

= (3n² - n) - (3(n - 1)² - (n - 1))

= (3n² - n) - (3n² + 3 - 6n - n + 1)

= 6n - 4

(ii) Since a_n = 6n - 4

∴ For first term, n = 1

By putting n = 1 in the nt^h term, we get,

a₁ = 6(1) - 4

= 2

∴ a = 2

(iii) Put n = 2 in the nth term, we get

a₂ = 6 × (2) - 4

= 12 - 4

= 8

Now common difference = d = a₂ - a₁

= 8 - 2

= 6

∴ Common difference = 6