Question

If S_n, the sum of first n terms of an AP is given by Sn = 3n² - 4n, find the nth term.

Answer 1

We have, S_n = 3n² - 4n
for n = 1 ⇒ `S_1 = 3 (1)^2 - 4(1) = 3 - 4 = -1`
⇒ a₁ = S1 = -1
for n = 2 ⇒ `S_2 = 3(2)^2 - 4(2)`
= 12 - 8 = 4
`S_2 = 4`
a₂ = S₂ - S₂
= 4 -(-1)
a₂ = 5

Common difference, d = a₂ - a₁
d = 5 -(-1)
d = 6
⇒ a_n = a + (n -1)d
= (-1) + (n -1)6
= -1 + 6n - 6
a_n = 6n - 7
⇒ n^th term is (6n - 7)