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 __

Question

Let $X$ be the set consisting of the first 2018 terms of the arithmetic progression $1, 6, 11, \dots,$ and Y be the set consisting of the first 2018 terms of the arithmetic progression $9, 16, 23, \dots$ . Then, the number of elements in the set $X \cup Y$ is _______. [2018]

Smart Achievers · Accepted Answer

(3748)

The given sequences upto 2018 terms are

$1, 6, 11, 16, \dots, 10086 and 9, 16, 23, \dots, 14128$

The common terms are

$16, 51, 86, \dots upto n terms, where T_{n} \leq 10086$

$\Rightarrow 16 + (n - 1) 35 \leq 10086$

$\Rightarrow 35 n - 19 \leq 10086$

$\Rightarrow n \leq \frac{10105}{35} = 288.7$ $∴$ $n = 288$

$∴$ $n (X \cup Y) = n (X) + n (Y) - n (X \cap Y)$

$= 2018 + 2018 - 288 = 3748$

Solution

Q.
Let $X$ be the set consisting of the first 2018 terms of the arithmetic progression $1, 6, 11, \dots,$ and Y be the set consisting of the first 2018 terms of the arithmetic progression $9, 16, 23, \dots$ . Then, the number of elements in the set $X \cup Y$ is _______. [2018]

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Solution

Q. 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 _______. [2018]

Ans. (3748) The given sequences upto 2018 terms are 1,6,11,16,…,10086 and 9,16,23,…,14128 The common terms are 16,51,86,… upto n terms, where Tn≤10086 ⇒16+(n-1)35≤10086 ⇒35n-19≤10086 ⇒n≤1010535=288.7 ∴n=288 ∴ n(X∪Y)=n(X)+n(Y)-n(X∩Y) =2018+2018-288=3748

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Q.
Let $X$ be the set consisting of the first 2018 terms of the arithmetic progression $1, 6, 11, \dots,$ and Y be the set consisting of the first 2018 terms of the arithmetic progression $9, 16, 23, \dots$ . Then, the number of elements in the set $X \cup Y$ is _______. [2018]