Let X be the set consisting of the first 2018 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$X=1,6,11,16,21,26,31,36,41,46,51,56,61,66,71,76,81,86, \ldots . .10086$

$Y=9,16,23,30,37,44,51,58,65,72,79,86, \ldots \ldots 14128$

$X \cap Y=1

Vedclass · Accepted Answer

$3748$

Vedclass · Answer

$X=1,6,11,16,21,26,31,36,41,46,51,56,61,66,71,76,81,86, \ldots . .10086$

$Y=9,16,23,30,37,44,51,58,65,72,79,86, \ldots \ldots 14128$

$X \cap Y=16,51,86,121, \ldots \ldots 	ext { (A.P. } d=35, a=16)$

So, $n(X \cap Y)=t$

$	ext { So, } 16+( t -1) 	imes 35 \leq 10086$

$t \leq 288.7$

$t =288$

$n(X \cup Y)=n(X)+n(Y)-n(X \cap Y)$

$=2018+2018-288$

$=3748$

Let $X$ be the set consisting of the first $2018$ terms of the arithmetic progression $1,6,11$,

. . . .and $Y$ be 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. . . .

Similar Questions

The difference between an integer and its cube is divisible by

If the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$ are in $A.P.$, then their common difference will be

If $\tan \,n\theta = \tan m\theta $, then the different values of $\theta $ will be in

If $\tan \left(\frac{\pi}{9}\right), x, \tan \left(\frac{7 \pi}{18}\right)$ are in arithmetic progression and $\tan \left(\frac{\pi}{9}\right), y, \tan \left(\frac{5 \pi}{18}\right)$ are also in arithmetic progression, then $|x-2 y|$ is equal to:

The sum of the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + ........$ to $9$ terms is

Let $X$ be the set consisting of the first $2018$ terms of the arithmetic progression $1,6,11$, . . . .and $Y$ be 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. . . .