a = 3, d = 15 – 3 = 12

Let last term be a_n

a_n = a + (n – 1) d

= 3 + (n – 1) 12

= 12n – 9 (i)

a₂₁ = a + 20d

= 3 + 20(12) = 243

Now, the term is 120 more than the 21^st term

a_n = 120 + a₂₁

= 120 + 243

= 363

Putting this value in (i), we get

363 = 12n – 9

12n = 363 + 9

n = 31

Hence, 31^st term of given A.P. is 120 more than its 21^st term