First term, a = 9, d = 12 – 9 = 3

Let, last term be a_n

a_n = a + (n – 1) d

= 9 + (n – 1)3

= 9 + 3n – 3 = 6 + 3n (i)

Let, 36^th term, a₃₆ = a + 35d

= 9 + 35 (3) = 114

Now the term is 39 more than its 36^th term

a_n = 39 + a₃₆

= 39 + 114 = 153

Putting the value in (i), we get

153 = 6 + 3N

3n = 153 – 6

3n = 147

n = 49

Hence, 49^th term of the given A.P. is 39 more than its 36^th term