Suppose that the number of terms in an $A.P.$ is $2k$,where $k \in N$. If the sum of all odd-positioned terms of the $A.P.$ is $40$,the sum of all even-positioned terms is $55$,and the last term of the $A.P.$ exceeds the first term by $27$,then $k$ is equal to:

Let the sum of $n, 2n, 3n$ terms of an $A.P.$ be $S_{1}, S_{2}$ and $S_{3}$ respectively. Show that $S_{3} = 3(S_{2} - S_{1})$.

If $S_n$ denotes the sum of $n$ terms of an arithmetic progression,then the value of $(S_{2n} - S_n)$ is equal to

Find the sum of odd integers from $1$ to $2001$.

If $a$ and $b$ are two numbers,$A$ is the arithmetic mean,and $S$ is the sum of $n$ arithmetic means between $a$ and $b$,then what does $S/A$ depend on?

If $a_r > 0, r \in N$ and $a_1, a_2, a_3, ..., a_{2n}$ are in an arithmetic progression,then $\frac{a_1 + a_{2n}}{\sqrt{a_1} + \sqrt{a_2}} + \frac{a_2 + a_{2n-1}}{\sqrt{a_2} + \sqrt{a_3}} + \frac{a_3 + a_{2n-2}}{\sqrt{a_3} + \sqrt{a_4}} + ... + \frac{a_n + a_{n+1}}{\sqrt{a_n} + \sqrt{a_{n+1}}} = ?$