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

Let the first term of the AP = a and common difference = d
Third term = 16
⇒ a₃ = 16 ⇒ a + 2d = 16 … (1)
7th term exceeds the 5th term by 12, therefore a₇ = a₅ + 12
⇒ a + 6d = a + 4d + 12
⇒ 2d = 12 ⇒ d = 6
Putting the value of d in equation (1), we get a + 2(6) = 16
⇒ a = 4
Hence, the A.P = a, a + 2d, … = 4, 10, 16, …