An AP consists of 50 terms of which 3rd term is 12 and the last term is 106.

Given, a₃ = 12 and a₅₀ = 106

a₃ = a + 2d = 12

a₅₀ = a + 49d = 106

Subtracting 3rd term from 50th term, we get;

a + 49d – a – 2d = 106 – 12

47d = 94

d = 2

Substituting the value of d in 12th term, we get;

a + 2 x 2 = 12

Or, a + 4 = 12

Or, a = 8

Now, 29th term can be calculated as follows:

a₂₉ = a + 28d

= 8 + 28 x 2

= 8 + 56 = 64