It is also given that $a_3=12$  and $a_{50}=106$      
Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of AP, we get
$a_{50}=a+(50-1)d\Rightarrow 106=a+49d\cdots \left(1\right)$  
$a_3=a+(3-1)d\Rightarrow 12=a+2d\cdots \left(2\right)$
Subtracting (1) and (2), we get,
$\Rightarrow 47d=94$
$\Rightarrow d=2$
Substituting $d$ in (2), we get,
$a+2\left(2\right)=12$
$a=12-4=8$

The ${29}^{th}$ term is,
$a_{29}=a+(29-1)d=8+28\left(2\right)=8+56=64$