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

Given a3 = 16 and a7 – a5 = 12

a3 = a + 2d = 16

a5 = a + 4d

a7 = a + 6d

As per question;

a + 6d – a – 4d = 12

⇒ 2d = 12

⇒ d = 6

Substituting the value of d in third term, we get;

a + 2 × 6 = 16

Or, a + 12 = 16

Or, a = 16 – 12 = 4

Thus, the AP can be given as follows:

4, 10, 16, 22, 28, …

18