Question

If s_n, the sum of first n terms of an AP is given by s_n= (3n²-4n), then find its nth term.

Answer 1

Let the sum of first n terms of the A.P.=S_n

Given: S_n = 3n² – 4n ...(i)

Now,

Replacing n by (n –1) in (i), we get,

S_{n – 1} = 3(n – 1)² – 4(n – 1)

n^th term of the A.P. a_n = S_n – S_{n – 1}

∴ a_n = (3n² – 4n) – [3(n – 1)² – 4(n – 1)]

⇒ a_n = 3 [ n² – (n – 1)²] – 4 [n – (n – 1)]

⇒ a_n = 3 (n² – n² + 2n – 1) – 4 (n – n + 1)

⇒ a_n = 3(2n –1) – 4

⇒ a_n = 6n – 3 – 4

⇒ a_n= 6n – 7

Thus, the n^th term of the A.P = 6n – 7.