a = 8, d = 6,

Last term = a_n

a_n= a + (n – 1) d

= 8 + (n – 1) 6

= 2 + 6n (i)

Let, 41^st term, a₄₁ = 8 + (41 – 1) 6

= 8 + 40 * 6 = 248

Now the term is 72 more than its 41^st term

a_n = 72 + a₄₁

= 72 + 248 = 320

Putting this value in (i), we get

320 = 2 + 6n

6n = 318

n = 53

Hence, 53^rd term of the given A.P. is 72 more than its 41^st term