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8. Sequences and Series
medium
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. . . .
A
$3747$
B
$3748$
C
$3749$
D
$3750$
(IIT-2018)
Solution
$X=1,6,11,16,21,26,31,36,41,46,51,56,61,66,71,76,81,86, \ldots . .10086$
$Y=9,16,23,30,37,44,51,58,65,72,79,86, \ldots \ldots 14128$
$X \cap Y=16,51,86,121, \ldots \ldots \text { (A.P. } d=35, a=16)$
So, $n(X \cap Y)=t$
$\text { So, } 16+( t -1) \times 35 \leq 10086$
$t \leq 288.7$
$t =288$
$n(X \cup Y)=n(X)+n(Y)-n(X \cap Y)$
$=2018+2018-288$
$=3748$
Standard 11
Mathematics
