Question

If the sum of the first n terms of an A.P. is given by S_n = (3n²- n), find its

(i) first term

(ii) common difference

(iii) nth term.

Answer 1

S_n = 3n² – n

Taking n = 1, we get

S₁ = 3(1)² - (1)

⇒ S₁ = 3 – 1

⇒ S₁ = 2

⇒ a₁ = 2

Taking n = 2, we get

S₂ = 3(2)² – 2

⇒ S₂ = 12 – 2

⇒ S₂ = 10

∴ a₂ = S₂ – S₁ = 10 – 2 = 8

Taking n = 3, we get

S₃ = 3(3)² – 3

⇒ S₃ = 27 – 3

⇒ S₃ = 24

∴ a₃ = 24 – 10 = 14

So, a = 1,

d = a₂ – a₁ = 8 - 2 = 6

Now, we have to find the 15^th term

a_n = a + (n – 1)d

a_n = 2 + (n – 1)6

a_n = 2 + 6n – 6

a_n = - 4 + 6n

Hence, the n^th term is 4n - 3.