$\begin{array}{1 1}9030 \\ 8127 \\ 8428 \\ 8729 \end{array} $

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- $n^{th}$ term of an A.P.$=t_n=a+(n-1)d$
- sum of $n$ terms of an A.P.$=\large\frac{n}{2}$$(l+a)$ where $l=t_n$

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$

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