Let $A=\left\{n\in N\mid n^2\le n+10,000\right\}, B=\left\{3k+1\mid k\in N\right\}$ and $C=\left\{2k\mid k\in N\right\},$ then the sum of all the elements of the set $A\cap \left(B-C\right)$ is equal to ________.

We have, $A=\left\{n\in \mathit{N}\mid n^2\le n+10,000\right\},B=\left\{3k+1\mid k\in \mathit{N}\right\}$ and $C=\left\{2k\mid k\in N\right\},$

$B\equiv \left\{4,7,10,13,16,19,\dots \dots \right\}$

$C\equiv \left\{2,4,6,8,10,12,14,16,20,\dots \dots \right\}$

$B-C\equiv \left\{7,13,19,\dots 97,\dots \right\}$

Now, $n^2-n\le 100\times 100$

$\Rightarrow n\left(n-1\right)\le 100\times 100$

$\Rightarrow A=\left\{1,2,\dots ,100\right\}$

So, $A\cap \left(B-C\right)=\left\{7,13,19,\dots ,97\right\}$

Let $S_n=7+13+19+\dots +97$

Clearly, the terms are in arithmetic progression. Here, $a=7$ and $d=13-7=6$ and $l=97$

$\Rightarrow l=a+\left(n-1\right)d=97$ where $n$ is the number of terms.

$\Rightarrow 7+\left(n-1\right)6=97$

$\Rightarrow \left(n-1\right)6=90$

$\Rightarrow \left(n-1\right)=15$

$\Rightarrow n=16$

Hence, sum $S_n=7+13+19+\dots +97$

$S_n=\frac{n}{2}\left(a+l\right)$

$S_n=\frac{16}{2}\left(7+97\right)$

$=832.$