Question

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 _____.

Answer 1

given : x be the set consisting of the first 2018 terms of the ap ; 1, 6, 11, .... and y be the set consisting of the first 2018 terms of the ap ; 9, 16, 23, .....

To find : The number of elements in the set (x U y)

solution : 1st ap ; 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, ..... [1 + (2018 - 1) × 5]

2nd ap ; 9, 16, 23, 30, 37, 44, 51, 58, 65 .... [9 + (2018 - 1) × 7]

common ap ; 16, 51 ..... r

here first term, a = 16

common difference, d = 51 - 16 = 35

so, rth term = 16 + (r - 1)35 ≤ 1 + (2018 - 1)5

⇒16 + (r - 1)35 ≤ 10086

⇒r ≤ 288.71

so, r = (x n y) = 288

now (x U y) = x + y - (y n y).

= 2018 + 2018 - 288 = 3748

Therefore the no of elements in the set of (x U y) is 3748

Answer 2

Answer:

3748

Step-by-step explanation:

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