Find the value of $\frac{\sqrt{a}+\sqrt{a+nd}}{\sqrt{a}+\sqrt{a+d}}+\frac{\sqrt{a}+\sqrt{a+nd}}{\sqrt{a+d}+\sqrt{a+2d}}+\dots +\frac{\sqrt{a}+\sqrt{a+nd}}{\sqrt{a+(n-1)d}+\sqrt{a+nd}}$

The correct option is D $n$

Given $\frac{\sqrt{a}+\sqrt{a+nd}}{\sqrt{a}+\sqrt{a+d}}+\frac{\sqrt{a}+\sqrt{a+nd}}{\sqrt{a+d}+\sqrt{a+2d}}+...+\frac{\sqrt{a}+\sqrt{a+nd}}{\sqrt{a+(n-1)d}+\sqrt{a+nd}}$

Raltionalize the denominators

$=(\sqrt{a}+\sqrt{a+nd})(\frac{1}{\sqrt{a}+\sqrt{a+d}}+\frac{1}{\sqrt{a+d}+\sqrt{a+2d}}+...+\frac{1}{\sqrt{a}+(n-1)d+\sqrt{a+2d}})$

$=(\sqrt{a}+\sqrt{a+nd})(\frac{1}{\sqrt{a}+\sqrt{a+d}}\times \frac{\sqrt{a+d}-\sqrt{a}}{\sqrt{a+d}-\sqrt{a}}+\frac{1}{\sqrt{a+d}+\sqrt{a+2d}}\times \frac{\sqrt{a+2d}-\sqrt{a+d}}{\sqrt{a+2d}-\sqrt{a+d}}$

$+...+\frac{1}{\sqrt{a+(n-1)d+\sqrt{a+nd}}}\times \frac{\sqrt{a+nd}-\sqrt{a+(n-1)d}}{\sqrt{a+nd}-\sqrt{a+(n-1)d}})$

$=(\sqrt{a}+\sqrt{a+nd})(\frac{\sqrt{a+d}-\sqrt{a}}{a+d-a}+\frac{\sqrt{a+2d}-\sqrt{a+d}}{a+2d-(a+d)}+\frac{\sqrt{a+3d}-\sqrt{a+2d}}{a+3d-(a+2d)}$

$+...+\frac{\sqrt{a+nd}-\sqrt{a+(n-1)d}}{a+nd-(a+(n-1\left)d\right)})$

$=(\sqrt{a}+\sqrt{a+nd})(\frac{\sqrt{a+d}-\sqrt{a}}{d}+\frac{\sqrt{a+2d}-\sqrt{a+d}}{d}+\frac{\sqrt{a+3d}-\sqrt{a+2d}}{d}$

$+...+\frac{\sqrt{a+nd}-\sqrt{a+(n-1)d}}{d})$

$=(\sqrt{a}+\sqrt{a+nd})(\frac{\sqrt{a+d}-\sqrt{a}+\sqrt{a+2d}-\sqrt{a+d}+\sqrt{a+3d}-\sqrt{a+2d}}{d}$

$+....+\frac{\sqrt{a+nd}-\sqrt{a+(n-1)d}}{d})$

$=\frac{(\sqrt{a}+\sqrt{a+nd})(\sqrt{a+nd}-\sqrt{a})}{d}$

$=\frac{(a+nd-a)}{d}=n$