a = 3, d = 15 – 3 = 12
Let last term be an
an = 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 21st term
an = 120 + a21
= 120 + 243
= 363
Putting this value in (i), we get
363 = 12n – 9
12n = 363 + 9
n = 31
Hence, 31st term of given A.P. is 120 more than its 21st term