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

To Find: A.P
Given: a₃ = 16, a₇ - a₅ = 12
For this question,
We need to find third, fifth and seventh term of an AP by formula of nth term.
We know that, nth term of an AP is given by :
a_n = a + (n - 1) d
So,
Given a₃ = 16 and a₇ – a₅ = 12
where a₃ = third term of the AP, and so on

a₃ = a + 2d = 16         ...........................eq(i)

a₅ = a + 4d

a₇ = a + 6d

As per question;
7th term exceeds the fifth term by 12, So the difference of seventh and fifth term will be 12

a + 6d – a – 4d = 12

⇒ 2d = 12

⇒ d = 6          

Substituting the value of d in eq (i), we get;

a + 2 × 6 = 16

Or, a + 12 = 16

Or, a = 16 – 12 = 4

Thus, the AP can be given as follows:
a, a+d, a+2d, a+3d, a+4d..... and thus,

4, 10, 16, 22, 28, …