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. . . . 

Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms.

Let $\mathrm{A}_{\mathrm{k}}=\mathrm{a}_1{ }^2-\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2-\mathrm{a}_4{ }^2+\ldots+\mathrm{a}_{2 \mathrm{k}-1}{ }^2-\mathrm{a}_{2 \mathrm{k}}{ }^2$.

If $\mathrm{A}_3=-153, \mathrm{~A}_5=-435$ and $\mathrm{a}_1{ }^2+\mathrm{a}_2{ }^2+\mathrm{a}_3{ }^2=66$, then $\mathrm{a}_{17}-\mathrm{A}_7$ is equal to....................

The sum of all natural numbers between $1$ and $100$ which are multiples of $3$ is

Let ${a_1},{a_2},\;.\;.\;.\;.,{a_{49}}$ be in $A.P.$ such that $\mathop \sum \limits_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m,$ then $m = \;\;..\;.\;.\;.\;$

Suppose $a_{1}, a_{2}, \ldots, a_{ n }, \ldots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms of the sum of first nine terms of the progression is $5: 17$ and $110< a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to -

Find the sum of integers from $1$ to $100$ that are divisible by $2$ or $5.$