jmc

Let $a$ be the smallest common term. We know that $$\begin{array}{rl}a& \equiv 1(\mathrm{mod}6)\\ a& \equiv 4(\mathrm{mod}7)\end{array}$$ We see that $a\equiv 1(\mathrm{mod}6)$ means that there exists a non-negative integer $n$ such that $a=1+6n$. Substituting this into $a\equiv 4(\mathrm{mod}7)$ yields $$1+6n\equiv 4(\mathrm{mod}7)⟹n\equiv 4(\mathrm{mod}7)$$ So $n$ has a lower bound of $4$. Then $n\ge 4⟹a=1+6n\ge 25$. We see that $25$ satisfies both congruences so $a=25$. If $b$ is any common term, subtracting $25$ from both sides of both congruences gives $$\begin{array}{rl}b-25& \equiv -24\equiv 0(\mathrm{mod}6)\\ b-25& \equiv -21\equiv 0(\mathrm{mod}7)\end{array}$$ Since $\gcd (6,7)=1$, we have $b-25\equiv 0(\mathrm{mod}6\cdot 7)$, that is, $b\equiv 25(\mathrm{mod}42).$ So $b=25+42m$ for some integer $m$. The largest such number less than $100$ is $\boxed{67}$, which happens to satisfies the original congruences.