**Solution:**

Given,

11^{th} term of an AP = a_{11} = 38

and 16^{th} term, a_{16} = 73

We know that,

a_{n} = a + (n − 1)d

a_{11} = a + (11 − 1)d

38 = a + 10d ….(1)

In the same way,

a _{16} = a + (16 − 1) d

73 = a + 15d ….(2)

On subtracting equation (1) from (2), we get;

35 = 5 d

d = 7

From equation (1),

38 = a + 10 × (7)

38 − 70 = a

a = −32

a_{31} = a + (31 − 1) d

= −32 + 30 (7)

= −32 + 210

= 178

Hence, the 31^{st} term of the AP is 178.