If $x_1,x_2,x_3,....,x_n$ are in Arithmetic Progression, then prove the following If $x_1,x_2,x_3,....,x_n$ are in Arithmetic Progression such that $x_i>0$ for all $i$, then prove that:
$$\frac{n-1}{\sqrt{x_1}+\sqrt{x_n}}-\frac{1}{\sqrt{x_{n-1}}+\sqrt{x_n}}=\frac{1}{\sqrt{x_1}+\sqrt{x_2}}+\frac{1}{\sqrt{x_2}+\sqrt{x_3}}+....+\frac{1}{\sqrt{x_{n-2}}+\sqrt{x_{n-1}}}$$
Could someone help me with this. I am not able to initiate this problem.
 A: Hint: If $h$ is the commum difference of the terms (ie $x_{k+1}-x_k=h$ for $k=1,\cdots n-1$) you have
$$\frac{1}{\sqrt{x_{k+1}}+\sqrt{x_k}}=\frac{\sqrt{x_{k+1}}-\sqrt{x_k}}{h}$$ 
A: Hint:
Let Arithmetic Progression is $x_1, x_2=x_1+d,x_3=x_2+d,..$
$$\frac1{\sqrt{x_i+1}+\sqrt{x_{i+2}}}=\frac{(\sqrt{x_{i+2}}-\sqrt{x_{i+1}})}{(\sqrt{x_{i+1}}+\sqrt{x_{i+2}})((\sqrt{x_{i+2}}-\sqrt{x_{i+1}}))}=\frac{\sqrt{x_{i+2}}-\sqrt{x_{i+1}}}{x_{i+2}-x_{i+1}}=\frac{\sqrt{x_{i+2}}-\sqrt{x_{i+1}}}{d}$$
Then
$$\frac{1}{\sqrt{x_1}+\sqrt{x_2}}+\frac{1}{\sqrt{x_2}+\sqrt{x_3}}+....+\frac{1}{\sqrt{x_{n-2}}+\sqrt{x_{n-1}}}=\frac{\sqrt{x_2}-\sqrt{x_1}+\sqrt{x_3}-\sqrt{x_2}+...+\sqrt{x_{n-1}}-\sqrt{x_{n-2}}}{d}=$$
$$=\frac{\sqrt{x_{n-2}}-\sqrt{x_1}}{d}$$
A: We have to evaluate the sum $\frac{1}{\sqrt{x_1}+\sqrt{x_2}}+\frac{1}{\sqrt{x_2}+\sqrt{x_3}}+....+\frac{1}{\sqrt{x_{n-2}}+\sqrt{x_{n-1}}}$.
To do this conjugate all of the roots.
So if $d$ is the common difference then  this sum is equal to the sum:
$\frac{\sqrt{x_2}-\sqrt{x_1}}{x_2-x_1}+\frac{\sqrt{x_3}-\sqrt{x_2}}{x_3-x_2}+....+\frac{\sqrt{x_{n-1}}-\sqrt{x_{n-2}}}{x_{n-1}-x_{n-2}}=\frac{\sqrt{x_2}-\sqrt{x_1}}{d}+\frac{\sqrt{x_3}-\sqrt{x_2}}{d}+....+\frac{\sqrt{x_{n-1}}-\sqrt{x_{n-2}}}{d}$
The sum is telescoping and equal to:
$\frac{\sqrt{x_{n-1}}-\sqrt{x_1}}{d}=\frac{\sqrt{x_{n-1}}-\sqrt{x_1}}{(x_n-x_1)/(n-1)}=(n-1)\frac{\sqrt{x_{n-1}}-\sqrt{x_1}}{x_n-x_1}=(n-1)(\frac{\sqrt{x_n}-\sqrt{x_1}}{x_n-x_1}-\frac{\sqrt{x_n}-\sqrt{x_{n-1}}}{x_n-x_1})=(n-1)(\frac{1}{\sqrt{x_1}+\sqrt{x_n}}-\frac{x_n-x_{n-1}}{(n-1)(x_n-x_{n-1})})=\frac{n-1}{\sqrt{x_1}+\sqrt{x_n}}-\frac{1}{\sqrt{x_{n-1}}+\sqrt{x_{n}}}$
