Determine the $AP$ whose third term is $16$ and the $7^{th}$ term exceeds the $5^{th}$ term by $12$.

(A) Let the first term of the $AP$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $AP$ is given by $a_n = a + (n - 1)d$.
Given that the third term is $16$:
$a_3 = a + (3 - 1)d = 16$
$a + 2d = 16$ --- $(1)$
Given that the $7^{th}$ term exceeds the $5^{th}$ term by $12$:
$a_7 - a_5 = 12$
$(a + 6d) - (a + 4d) = 12$
$2d = 12$
$d = 6$
Substituting $d = 6$ in equation $(1)$:
$a + 2(6) = 16$
$a + 12 = 16$
$a = 4$
Thus,the $AP$ is $a, a+d, a+2d, a+3d, \ldots$
$4, 4+6, 4+12, 4+18, \ldots$
$4, 10, 16, 22, \ldots$