The number of common terms to the two sequence $17,21,25,\dots \dots ,417\text{ and }16,21,26,\dots ..,466$ is :

In $17,21,25,\dots ,417$, number of terms= 101. and $T_n=17+(n-1)4$

In $16,21,26,\dots ,466$, number of terms= 91 and $T_m=16+(m-1)5$

$\begin{array}{l}{T}_n=T_m\\ \Rightarrow 17+(n-1)4=16+(m-1)5,1\le n\le 101\end{array}$

$1\le m\le 91$

$\therefore 5m-4n=2n=m=2$ is a solution. 

Then $n=2,7,\dots ;2+(k-1)5\le 101\Rightarrow k=20$

Also $m=2,6,10,\dots$

Hence, there are 20 common terms.