**Solution:**

Given,

The third term of an AP, a_{3} = 16

As we know,

a + (3 − 1) d = 16

a + 2d = 16 ………………………………………. (i)

It is given that, the 7^{th} term exceeds the 5^{th} term by 12.

a_{7} − a_{5} = 12

[a + (7 − 1) d ] − [a + (5 − 1)d ] = 12

(a + 6d) − (a + 4d) = 12

2d = 12

d = 6

From equation (i), we get,

a + 2(6) = 16

a + 12 = 16

a = 4

Therefore, AP will be 4, 10, 16, 22, …