In the series $3,7,11,15,\mathrm{.......}$  and $2,5,8,\mathrm{.......}$  each continued to $100\text{terms}$ , the number of terms that are identical is

Let the $n^{th}$  term of the first series $=$  the $m^{th}$  term of the second series

$3+\left(n-1\right)4=2+\left(m-1\right)3$

$3+4n-4=2+3m-3$

$4n-1=3m-1$

$4n=3m$

$\frac{n}{3}=\frac{m}{4}=k$

$n=3k$  (or) $m=4k$

As each series is continued to $100\text{terms}$

$n=3k\le 100$  and $m=4k\le 100$

Possible values of $k$  are $12,3,\mathrm{.....25}$  and corresponding to each value of $k$

$\therefore$  There are $25$  identical terms.