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Let $A$ be the set consisting of the first $2020$ terms of the arithmetic progression $1,3, 5,7,9,\ldots,$ and $B$ be the set consisting of the first $2020$ terms of the arithmetic progression $5,10,15,20,25,30,\ldots $. The number of elements in the set $A\cup B$ is ________
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$n(A\cup B) = n(A)+n(B)- n(A\cap B)$

$n(A) = 2020$

So, $t_n = t_1 + (n-1)d,$ where $t_1 = 1,n = 2020$ and $d = 2$

$\implies t_n = 1 +2019\times 2 = 4039$

$\implies A=1,3,5,7,9,11,13,15,17,19,21,\dots, 4039$

$n(B) = 2020$

So, $t_n = t_1 + (n-1)d,$ where $t_1 = 5,n = 2020$ and $d = 5$

$\implies t_n = 5 +2019\times 5 = 10100$

$\implies B=5,10,15,20,25,30,\dots ,10100$

$C=A\cap B = 5,15,25,\dots, 4039 $

$5+(n-1)10\leq 4039$

$10n-5\leq 4039$

$10n\leq 4044$

$n\leq 404.4$

$n=404$

$n(C) = n(A\cap B) = 404$

Therefore, $n(A\cup B) = 2020+2020-404 = 3636$

So, the correct answer is $3636$.
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