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

Let $a =$ first term, $d =$ common difference

Third term: $a_3 = a + 2d = 16$ ...(i)

Also, $a_7 - a_5 = 12$

$$(a + 6d) - (a + 4d) = 12$$

$$2d = 12$$

$$d = 6$$

Put $d = 6$ in (i):

$$\begin{aligned} a + 2(6) &= 16 \\ a + 12 &= 16 \\ a &= 4 \end{aligned}$$

Therefore, the AP is 4, 10, 16, 22, 28, ...