Mark the correct alternative in the following question:

As, the common difference of the A.P. $3,7,11,15, \ldots=7-3=4$ and

the common difference of the A.P. $1,6,11,16, \ldots=6-1=5$

And, the common terms of both the A.P.s will be in A.P.

So, the common difference of the A.P. of the common terms, $d=\operatorname{LCM}(4,5)=4 \times 5=20$ and

its first common term, $a=11$

Now, the tenth common term, $a_{10}=a+(10-1) d$

$=11+9 \times 20$

$=11+180$

$=191$

Hence, the correct alternative is option (a).