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 sum of first $n$ natural numbers is

If $\log _{3} 2, \log _{3}\left(2^{x}-5\right), \log _{3}\left(2^{x}-\frac{7}{2}\right)$ are in an arithmetic progression, then the value of $x$ is equal to $.....$

If the sum of the first $n$ terms of a series be $5{n^2} + 2n$, then its second term is

Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=n(n+2)$

If ${A_1},\,{A_2}$ be two arithmetic means between $\frac{1}{3}$ and $\frac{1}{{24}}$ , then their values are

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. . . .