Question

Let X be a 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 ∪ Y is ____
(jee advanced 2019)​

Answer 1

Answer:

3748 is the required answer

Step-by-step explanation:

Let m=n(X∩Y)

∴ 16 + (m − 1) ×35 ≤ 10086

⇒m ≤ 288.71

⇒m = 288

∴n(X∪Y)=n(X)+n(Y)−n(X∩Y)

⇒2018 + 2018 − 288 = 3748

Answer 2

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X:1,6,11,_________, 10086

Y:9,16,23,________, 14128

X∩Y :16,51,86, _________

Let m=n(X∩Y)

∴16+(m−1)×35≤10086

⇒m≤288.71

⇒m=288

∴n(X∪Y)=n(X)+n(Y)−n(X∩Y)

=2018+2018−288=3748.