Let first term of AP is a and common difference of AP is d.

**We know that,**

n^{th} term of AP is an = a + (n – 1) d.

**Given that,**

8^{th} term of AP is a_{8} = 31

⇒ a + (8 – 1) d = 31

⇒ a + 7d = 31 **… (1) **

**Also,**

**Given that,**

15th term of AP is 16 more than the 11^{th} term of AP.

i.e., a_{15} = 16 + a_{11}

⇒ a + 14d = 16 + a + 10d

⇒ 14d − 10d = 16

⇒ d = \(\frac{16}{4}\) = 4.

**From equation (1), **

a + 7 × 4 = 31

**(∵d = 4) **

⇒ a = 31 – 28 = 3.

**Hence, **

The first term of AP is a_{1} = a = 3.

The second term of AP is a_{2} = a + d

= 3 + 4 = 7.

The third term of AP is a_{3} = a + 2d

= 3 + 8

= 11.

**Hence, **

The required AP is 3, 7, 11, ……… .