If $\sum_{r=1}^n(2r+1)=440$,then $n = \ldots$.

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

The $p^{\text{th}}$,$q^{\text{th}}$,and $r^{\text{th}}$ terms of an $A.P.$ are $a$,$b$,and $c$ respectively. Show that $(q-r)a + (r-p)b + (p-q)c = 0$.

Three positive numbers $a, b,$ and $c$ are in an Arithmetic Progression $(AP)$ and $abc = 4$. The minimum possible value of $b$ is:

If $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ are consecutive terms of an $A.P.$,then $(b - c)^2, (c - a)^2, (a - b)^2$ will be in

Insert five numbers between $8$ and $26$ such that the resulting sequence is an $A.P.$