JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 21)
Let A = {n $$\in$$ N | n2 $$\le$$ n + 10,000}, B = {3k + 1 | k$$\in$$ N} an dC = {2k | k$$\in$$N}, then the sum of all the elements of the set A $$\cap$$(B $$-$$ C) is equal to _____________.
Answer
832
Explanation
B $$-$$ C $$ \equiv $$ {7, 13, 19, ......, 97, .......}
Now, n2 $$-$$ n $$\le$$ 100 $$\times$$ 100
$$\Rightarrow$$ n(n $$-$$ 1) $$\le$$ 100 $$\times$$ 100
$$\Rightarrow$$ A = {1, 2, ......., 100}.
So, A$$\cap$$(B $$-$$ C) = {7, 13, 19, ......., 97}
Hence, sum = $${{16} \over 2}(7 + 97) = 832$$
Now, n2 $$-$$ n $$\le$$ 100 $$\times$$ 100
$$\Rightarrow$$ n(n $$-$$ 1) $$\le$$ 100 $$\times$$ 100
$$\Rightarrow$$ A = {1, 2, ......., 100}.
So, A$$\cap$$(B $$-$$ C) = {7, 13, 19, ......., 97}
Hence, sum = $${{16} \over 2}(7 + 97) = 832$$
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