Which term of the AP 3, 15, 27, 39, ……… will be 120 more than its 21^st term?

Or

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

Consider the AP 3, 15, 27, 39, ……

We know that the n^th term of an AP is given by:

a_n = a + (n – 1)d

a₂₁ = 3 + (21 – 1)12

⇒ a₂₁ = 3 + (20)12

⇒ a₂₁ = 3 + 240

⇒ a₂₁ = 243

According to the question,

a_n = 120 + a₂₁

⇒a + (n – 1)d = 120 + 243

⇒3 + (n – 1)12 = 363

⇒ 3 + 12n - 12 = 363

⇒ 12n - 9 = 363

⇒ 12n = 372

⇒ n = 31

So, 31^st term of the AP is 120 more than its 21^st term i.e. 363.

OR

S_n = 3n² – 4n

At n = 1, S₁ = -1

At n = 2, S₂ = 4

At n = 3, S₃ = 15

We know that an AP is also represented by sum of nth terms.

AP = S₁, (S₂ – S₁), (S₃ – S₂), ……

AP = -1,(4+1),(15-4), .......  

AP = -1, 5, 11, ……

a_n = a + (n – 1)d

a_n = -1 + (n – 1)6
= -1 + 6n - 6
= 6n – 7

So, nth term of an AP is given by 6n – 7.