The largest term common to the sequences 1, 11, 21, 31, ............. to 100 terms and 31, 36, 41, 46 ......... to 100 terms is

The given sequence is $1,11,21,31,...$

The first term of the sequence is $a_1=1$ and the common difference is $d=10$

The other sequence is $31,36,41,46,...$

The first term of the sequence is $A_1=31$ and the common difference is $D=5$

$\begin{array}{rcl}1+\left(n-1\right)10& =& 31+\left(m-1\right)5\\ 1+10n-10& =& 31+5m-5\\ 10n-5m& =& 35\\ 2n-m& =& 7\end{array}$

find the value of either $n$ or $m$ such that $2n-m=7$ and $n,m\le 100$

here $m$ must be odd number

suppose that $m=99$

        $\begin{array}{rcl}2n& =& 7+99\\ & =& 106\\ n& =& 53\end{array}$

Hence hte largest common term is $53rd$ term of the first sequence and 99th term of the second sequence

hence, $t_{53}=1+52\left(10\right)=521$

Therefore, the common term in both sequences is $\boxed{521}$