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
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Solution
n2−n−10000≤0 n∈(1−√400012,1+√400012)
But n∈N ∴A={1,2,3,.......,100} and B−C={3k+1|k∈even} ⇒B−C={7,13,19,.......,97,..} ∴A∩(B−C)={7,13,19,...,97}
Sum of all elements =162[7+97]=832