Question

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

Answer 1

Given AP is 

3, 15, 27, 39  .....

where a = 3, d = 15 - 3 = 12

Let the nth term be 120 more than its 21st term.

t_n = t₂₁ + 120

⇒ 3 +(n -1 )12 = 3 +20 × 12 + 120 

⇒ (n - 1) × 12 = 363 - 3 

⇒ (n - 1) =`360/12`

∴ n = 31 

Hence , the required term is 

t₃₁ = 3 + 30 × 12 

  = 363