An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution: $a_{50}=106$
a+(50-1)=106-(1)
$a_3=12$
a+2d=12-(2)
subtracting equation 2 from 1
47d=94
d=2
putting this value in equation 2
a=8
$a_{29}$=a+28d
=8+28(2)
=8+56
=64