What is the sum of all numbers between $100$ and $300$ that are divisible by $7$?

Let $a_n, n \geq 1$,be an arithmetic progression with first term $2$ and common difference $4$. Let $M_n$ be the average of the first $n$ terms. Then the sum $\sum_{n=1}^{10} M_n$ is

Find the sum of all natural numbers lying between $100$ and $1000$ which are multiples of $5.$

The $15^{th}$ term of the arithmetic progression $4 + 9 + 14 + 19 + \dots$ is......

If $a_{1}, a_{2}, a_{3}, \ldots$ and $b_{1}, b_{2}, b_{3}, \ldots$ are $A.P.$ and $a_{1}=2, a_{10}=3, a_{1}b_{1}=1=a_{10}b_{10}$,then $a_{4}b_{4}$ is equal to

If the sum of the first $11$ terms of an $A.P.$,$a_{1}, a_{2}, a_{3}, \ldots$ is $0$ $(a_{1} \neq 0)$,then the sum of the $A.P.$,$a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ is $k a_{1}$,where $k$ is equal to