Let $A$ be the set of first $101$ terms of an $A$.$P$.,whose first term is $1$ and the common difference is $5$, and let $B$ be the set of first $71$ terms of an $A$.$P$.,whose first term is $9$ and the common difference is $7$. Then, the number of elements in $A \cap B$ which are divisible by $3$ is:

For a real number $x$,$[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $\left[\frac{1}{2}\right] + \left[\frac{1}{2} + \frac{1}{100}\right] + \left[\frac{1}{2} + \frac{2}{100}\right] + \left[\frac{1}{2} + \frac{3}{100}\right] + \ldots + \left[\frac{1}{2} + \frac{99}{100}\right] = $

If $P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$ is a polynomial such that $P(0) = 1, P(1) = 2, P(2) = 5, P(3) = 10$ and $P(4) = 17$,then $P(5) =$

Let $x_k$ be real numbers such that $x_k \geq k^4+k^2+1$ for $1 \leq k \leq 2018$. Denote $N=\sum_{k=1}^{2018} k$. Consider the following inequalities.
$I$. $\left(\sum_{k=1}^{2018} k x_k\right)^2 \leq N\left(\sum_{k=1}^{2018} k x_k^2\right)$
$II$. $\left(\sum_{k=1}^{2018} k x_k\right)^2 \leq N\left(\sum_{k=1}^{2018} k^2 x_k^2\right)$
Then,

Let $A = \{ x \in R : [x + 3] + [x + 4] \leq 3 \}$ and $B = \{ x \in R : 3^x \left( \sum_{n=1}^{\infty} \frac{3}{10^n} \right)^{x-3} < 3^{-3x} \}$,where $[t]$ denotes the greatest integer function. Then,

If $U$ is the universal set and $A \cup B \cup C = U$,then ${(A - B) \cup (B - C) \cup (C - A)}'$ is equal to: