Let the larger number be x and the smaller number be y.
Then, we have:
x –y = 26 ………….(i)
x = 3y …………(ii)
On substituting x = 3y in (i), we get:
3y –y = 26 ⇒ 2y = 26 ⇒ y = 13
On substituting y = 13 in (i), we get:
x – 13 = 26 ⇒ x = 26 + 13 = 39
Hence, the required numbers are 39 and 13.