Subjective Type

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 _______?

Solution

X:1,6,11,X:1,6,11, _________, 1008610086 Y:9,16,23,Y:9,16,23, ________, 1412814128 X∩YX∩Y :16,51,86:16,51,86 , _________ Let m=n(X∩Y)m=n(X∩Y) ∴16+(m−1)×35≤10086∴16+(m−1)×35≤10086 ⇒m≤288.71⇒m≤288.71 ⇒m=288⇒m=288 ∴n(X∪Y)=n(X)+n(Y)−n(X∩Y)∴n(X∪Y)=n(X)+n(Y)−n(X∩Y) =2018+2018−288=3748=2018+2018−288=3748 .

SIMILAR QUESTIONS

Sequences and Series

Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than $$200$$ and less than $$220$$. If the second term in it is $$12$$, then $$4^{th}$$ term is :

Sequences and Series

$$\displaystyle { a }_{ n }=n(n+2)$$

Sequences and Series

$$\displaystyle { a }_{ n }=\frac { n }{ n+1 } $$

Sequences and Series

$$\displaystyle{ a }_{ n }={ 2 }^{ n }$$

Sequences and Series

$$\displaystyle{ a }_{ n }=\frac { 2n-3 }{ 6 } $$

Sequences and Series

$$\displaystyle{ a }_{ n }={ (-1) }^{ n-1 }\quad { 5 }^{ n+1 }$$

Sequences and Series

$$\displaystyle{ a }_{ n }=n\frac { { n }^{ 2 }+5 }{ 4 } $$

Sequences and Series

$$\displaystyle{ a }_{ n }=4n-3;{ a }_{ 17 },{ a }_{ 24 }$$

Sequences and Series

$$\displaystyle { a }_{ n }=\frac { { n }^{ 2 } }{ { 2 }^{ n } } ;{ a }_{ 7 }$$

Sequences and Series

$$\displaystyle{ a }_{ n }=({ -1 })^{ n-1 }{ n }^{ 3 };{ a }_{ 9 }$$

Contact Details