Let A= n ∈ N mid n2 ≤ n+10,000 B= 3 k+1 mid k ∈ N and C= 2 k mid k ∈ N then the sum of all the elements of the set A ∩(B-C) is equal to

Question

Let A= n ∈ N mid n2 ≤ n+10,000 B= 3 k+1 mid k ∈ N and C= 2 k mid k ∈ N then the sum of all the elements of the set A ∩(B-C) is equal to  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

B-C ≡ 7,13,19, ldots 97, ldots Now, n2-n ≤ 100 × 100 ⇒ n(n-1) ≤ 100 × 100 ⇒ A= 1,2, ldots ldots, 100 So, A ∩(B-C)= 7,13,19, ldots ldots, 97 Hence, sum =(16/2)(7+97)=832

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

$B - C \equiv {7, 13, 19, \dots 97, \dots}$
Now, $n^{2} - n \leq 100 \times 100$
$\Rightarrow n (n - 1) \leq 100 \times 100$
$\Rightarrow A = {1, 2, \dots\dots, 100}$
So, $A \cap (B - C) = {7, 13, 19, \dots\dots, 97}$
Hence, sum $= \frac{16}{2} (7 + 97) = 832$

Q. Let A={n∈N∣n2≤n+10,000}, B={3k+1∣k∈N} and C={2k∣k∈N}, then the sum of all the elements of the set A∩(B−C) is equal to ___

B−C≡{7,13,19,…97,…} Now, n2−n≤100×100 ⇒n(n−1)≤100×100 ⇒A={1,2,……,100} So, A∩(B−C)={7,13,19,……,97} Hence, sum =216​(7+97)=832

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

$B - C \equiv {7, 13, 19, \dots 97, \dots}$
Now, $n^{2} - n \leq 100 \times 100$
$\Rightarrow n (n - 1) \leq 100 \times 100$
$\Rightarrow A = {1, 2, \dots\dots, 100}$
So, $A \cap (B - C) = {7, 13, 19, \dots\dots, 97}$
Hence, sum $= \frac{16}{2} (7 + 97) = 832$