Numbers between $200$ and $400$ divisible by $7$ forms the sequence
$203,210,217........399$
This sequence is an A. P. with first term $a=203$ and
common difference $d=7$
The last term or $n^{th}$ term of this sequence is $t_n=399$
We know that $n^{th}$ term of an A.P. = $a+(n-1)d$
$\Rightarrow\:a+(n-1)d=399$
$\Rightarrow\:203+(n-1)7=399$
$\Rightarrow\:(n-1)7=399-203=196$
$\Rightarrow\:n-1=\large\frac{196}{7}$$=28$
or $n=29$
$\Rightarrow\:$The sum of the required numbers =
$203+210+217+......399=\large\frac{n}{2}$$(t_n+a)$
$=\large\frac{29}{2}$$(399+203)=29\times301=8729$