Let `X` be the set consisting of the first 2018 terms of the arithmetic progression `1,\ 6,\ 11 ,\ ddot,` and `Y` be the set consisting of the first 2018 terms of the arithmetic progression `9,\ 16 ,\ 23 ,\ ddot` . Then, the number of elements in the set `XuuY` is _____.

Answer

3748

Explanation

(3748)
X={1,6,11,16….2018 terms }
Y={9,16,23,….,2018 terms }
`n(x cup Y)=n(y)-n(X cap Y)`
For `x(x cup Y)` , we have to find common terms in two A.Ps Common terms will form an A.P with common difference as L.C.M of common differences of the given A.P.s
Common difference of set X is 5 and tha of set Y is 7
So, common difference of A.P of common terms is `5 xx 7 = 35`
So, common term are 16,51,86.
Now ,last terms are in set `X is 1+(2018 -1)xx 7 = 14128 `
so , there will be 288 common termes .
`therefore n(X cup Y) = n(Y)-n(X cap Y)`
= 2018+ 2018 -288
= 3748