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

Or, 2d = 12

Or, d = 6

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

a + 2 x 12 = 16

Or, a + 24 = 16

Or, a = 16 – 24 = - 8

Thus, the AP can be given as follows:

-8, 4, 16, 28, 40, …