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

Question

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

Professor PC Thomas Classes · Answer

(832) : We have,\ $A=\{1,2,3, \ldots \ldots, 100\} \quad\left(\because 101^{2}=10201>ight.$ 10101)\ $B=\{4,7,10, \ldots \ldots ., 100,103,106, \ldots \ldots$.\ $C=\{2,4,6,8, \ldots \ldots ., 100,102,104,106, \ldots \ldots$.\ Now, $B-C=\{7,13,19,25, \ldots \ldots \ldots ., 97,103$, .......}\ $\Rightarrow \quad A \cap(B-C)=\{7,13,19,25, \ldots \ldots . .97\}$\ We can see that the elements of the set $A \cap(B-C)$ form an A.P. with $a=7, d=6$.\ $	herefore \quad 97=7+(n-1) 6 \Rightarrow n=16$.\ So, $S_{16}=\frac{16}{2}[7+97]=832$\