Let the a be first term and d be common difference

As we know

a_n = a + (n – 1)d

Given,

⇒ a₈ = 31

⇒ a + 7d = 31

⇒ a = 31 – 7d …[1]

Also, As 15^th term is 16 more than 11^th term

⇒ a₁₅ = a₁₁ + 16

⇒ a + 14d = a + 10d + 16

⇒ 4d = 16

⇒ d = 4

Using this value in equation [1]

a = 31 – 7(4) = 3

So, AP is

a, a + d, a + 2d, …

3, 3 + 4, 3 + 2(4), …

3, 7, 11, …