**Solution:**

Given,

3^{rd} term of an AP = a_{3} = 12

And 50^{th} term, a_{50} = 106

We know that,

a_{n} = a + (n − 1) d

a_{3} = a + (3 − 1) d

12 = a + 2d ….(1)

In the same way,

a_{50 }= a + (50 − 1) d

106 = a + 49 d ….(2)

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

94 = 47 d

d = 2 = common difference

From equation (1),

12 = a + 2(2)

a = 12 − 4 = 8

a_{29} = a + (29 − 1) d

a_{29} = 8 + (28)2

a_{29} = 8 + 56 = 64

Therefore, the 29^{th} term is 64.