JEE Advanced2018MathematicsSequences and SeriesActual
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 ∪ Y is___.
Correct answer
0
Step-by-step solution
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