Let A = {n in N : n text{ is a } 3 text{-digi | vedclass

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(D) The set $A$ consists of all $3$-digit numbers,i.e.,$A = \{100, 101, \dots, 999\}$.$B = \{9k + 2 : k \in N\} \cap A = \{101, 110, \dots, 992\}$. Th

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$5$

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(D) The set $A$ consists of all $3$-digit numbers,i.e.,$A = \{100, 101, \dots, 999\}$.$B = \{9k + 2 : k \in N\} \cap A = \{101, 110, \dots, 992\}$. This is an arithmetic progression with $a = 101$,$d = 9$,and $l_n = 992$. The number of terms $n_B$ is given by $992 = 101 + (n_B - 1)9$ $\Rightarrow 891 = (n_B - 1)9$ $\Rightarrow n_B - 1 = 99$ $\Rightarrow n_B = 100$.Sum $S(B) = \frac{100}{2}(101 + 992) = 50 	imes 1093 = 54650$.Similarly,$C = \{9k + l : k \in N\} \cap A$. For $3$-digit numbers,$k$ ranges from $11$ to $110$ (since $9(11)+l = 99+l \ge 100$ and $9(110)+l = 990+l \le 999$ for $l Sum $S(C) = \sum_{k=11}^{110} (9k + l) = 9 	imes \frac{100}{2}(11 + 110) + 100l = 9 	imes 50 	imes 121 + 100l = 54450 + 100l$.Given $S(B \cup C) = S(B) + S(C) - S(B \cap C) = 109600$.If $l 
eq 2$,then $B \cap C = \phi$,so $S(B \cup C) = 54650 + 54450 + 100l = 109100 + 100l = 109600$.$100l = 500 \Rightarrow l = 5$.

Let $A = \{n \in N : n \text{ is a } 3 \text{-digit number}\}$. Let $B = \{9k + 2 : k \in N\}$ and $C = \{9k + l : k \in N\}$ for some $l$ $(0 < l < 9)$. If the sum of all the elements of the set $A \cap (B \cup C)$ is $274 \times 400$,then $l$ is equal to:

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