Let A = n ∈ N: n is a 3 - digit number B = \ 9k + 2: k ∈ N\ and C =...

$B$ and $C$ will contain three digit numbers of the form $9k + 2$ and $9k + l$ respectively. We need to find sum of all elements in the set $B \cup C$ effectively.

Now, $S$\left(B \cup C\right) = S$\left(B\right) + S\left(C\right) - S\left(B \cap C\right)$ where $S ( k )$ denotes sum of elements of set $k .$

Also, $B = \left\{101, 110, \ldots \ldots , 992\right\}$

$∴ S\left(B\right) = \dfrac{100}{2}\left(101 + 992\right) = 54650$

Case-I: If $l = 2$

then $B \cap C = B$

$∴ S\left(B \cup C\right) = S\left(B\right)$

which is not possible as given sum is

$274 \times 400 = 109600 .$

Case-II If $l \ne 2$

then $B \cap C = ϕ$

$∴ S\left(B \cup C\right) = S\left(B\right) + S\left(C\right) = 400 \times 274$

$⇒ 54650 + \mathop{∑}_{k = 1}^{110}9k + l = 109600$

$⇒ 9\mathop{∑}_{k = 11}^{110}k + \mathop{∑}_{k = 11}^{110}l = 54950$

$⇒ 9\left(\dfrac{100}{2}\left(11 + 110\right)\right) + l\left(100\right) = 54950$

$⇒ 54450 + 100l = 54950$

$⇒ l = 5$