JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 9)
Let X be the set consisting of the first 2018 terms of the arithmetic progression 1, 6, 11, ...., and Y be the set consisting of the first 2018 terms of the arithmetic progression 9, 16, 23, .... . Then, the number of elements in the set X $$ \cup $$ Y is .........
Answer
3748
Explanation
Here, X = {1, 6, 11, ....., 10086}
[$$ \because $$ an = a + (n $$-$$ 1)d]
and Y = {9, 16, 23, ..., 14128}
X $$ \cap $$ Y = {16, 51, 86, ...}
tn of X $$ \cap $$ Y is less than or equal to 10086
$$ \therefore $$ tn = 16 + (n $$-$$ 1) 35 $$ \le $$ 10086
$$ \Rightarrow $$ n $$ \le $$ 288.7
$$ \therefore $$ n = 288
$$ \because $$ n(X $$ \cap $$ Y) = n(X) + n(Y) $$-$$ n(X $$ \cap $$ Y)
$$ \therefore $$ n(X $$ \cap $$ Y) = 2018 + 2018 $$-$$ 288 = 3748
[$$ \because $$ an = a + (n $$-$$ 1)d]
and Y = {9, 16, 23, ..., 14128}
X $$ \cap $$ Y = {16, 51, 86, ...}
tn of X $$ \cap $$ Y is less than or equal to 10086
$$ \therefore $$ tn = 16 + (n $$-$$ 1) 35 $$ \le $$ 10086
$$ \Rightarrow $$ n $$ \le $$ 288.7
$$ \therefore $$ n = 288
$$ \because $$ n(X $$ \cap $$ Y) = n(X) + n(Y) $$-$$ n(X $$ \cap $$ Y)
$$ \therefore $$ n(X $$ \cap $$ Y) = 2018 + 2018 $$-$$ 288 = 3748
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