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, …