$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?

I was casting about for a method that didn't require a lot of work with multiple-angles and trigonometric identities. Up to now, I had a different -- though ultimately related -- argument (without using a circumscribed circle) which led me to the same equation, $ \  \frac{1}{\sin\theta} \ = \ \frac{1}{\sin2\theta} \ + \  \frac{1}{\sin3\theta}  $ , which Shailesh already produced.  Here's something using more basic triangle geometry than what I'd had.

For convenience, we will call the lengths of the sides of the polygon $ \ s \ $ , making $ \ A_1A_2 \ = \ s \ $ and call the other lengths of interest $ \ A_1A_3 \ = \ t \ $ and $ \ A_1A_4 \ = \ u \ $ .  
"Drop perpendiculars" to $ \ \overline{A_1A_4} \ $ from $ \ A_2 \ $ to produce point $ \ P \ $ and from $ \ A_3 \ $ to produce $ \ Q \ $ .  Since we are working with a regular polygon, it is straightforward to show that $ \ A_1A_2A_3A_4 \ $ is a trapezoid and that $ \ PQ \ = \ s \ $ .  Extend, say , $ \ A_4A_3 \ $ to a point $ \ R \ $ : since $ \ \angle A_2A_3R \ $ is an exterior angle of the polygon,   $ \ m(\angle A_2A_3R) \ = \ \frac{2 \pi}{n} \ $ . We have $ \ \overline{A_2A_3} \ \ \Vert \ \ \overline{A_1A_4} \ $ , so corresponding angle $ \ \angle QA_4A_3 \ $ also has measure $  \ \frac{2 \pi}{n} \ $ .
Consequently, $ \ A_4Q \ = \ A_1P \ = \ s \ \cos \left(\frac{2 \pi}{n} \right) \ $ [it is clear that the two segments are congruent] and so
$$ u \ = \ s \ \left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right) \ \  . $$ 
Although $  \ \Delta A_1A_2A_3 \ $ is isosceles, and   $ \ m(\angle A_1A_2A_3) \ = \ \pi \ - \ \frac{2 \pi}{n} \ $   by a familiar theorem (or because it is supplementary to an exterior angle), we will actually not exploit the Law of Cosines here to find $ \ t \ $ .  Instead, we obtain that $ \ m(\angle A_1A_3A_2 ) \ = \ \frac{1}{2} \ ( \ \pi \ - \ [ \pi \ - \ \frac{2 \pi}{n} ] \ ) \ = \ \frac{\pi}{n} \ $ . From this, we find $ \ m(\angle A_1A_3A_4 ) \ = \ \left(\pi \ - \ \frac{2 \pi}{n} \right) \ - \frac{\pi}{n} $ $ \ = \ \left(\pi \ - \ \frac{3 \pi}{n} \right) \ $ .
By the Law of Sines, 
$$ \frac{t}{\sin \left(  \frac{2 \pi}{n}  \right)} \ = \ \frac{u}{\sin \left( \ \pi \ - \ \frac{3 \pi}{n} \ \right) \ } \ \ \Rightarrow \ \ \frac{t}{\sin \left(  \frac{2 \pi}{n}  \right)} \ = \ \frac{s \ \left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)}{\sin \left(  \frac{3 \pi}{n} \right)  }  $$
[using the identity "sine of an angle equals the sine of its supplement"].
At last coming to the equation under discussion, we obtain
$$ \frac{1}{s} \ = \ \frac{1}{t} \ + \ \frac{1}{u} \ = \ \frac{1}{s} \ \left[ \ \frac{1 }{\left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)} \ \left( \ \frac{\sin \left(   \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right) \ \right] \ \left[ \ 1 \ + \ \left( \ \frac{\sin \left(   \frac{2 \pi}{n} \right)}{\sin \left( \ \frac{3 \pi}{n} \ \right)} \ \right) \ \right] $$
$$ \Rightarrow \ \ 1 \ = \ \frac{1 }{\left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)} \ \ \left[ \   \left( \ \frac{\sin \left(   \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right) \ + \ 1 \ \right] $$
$$ \Rightarrow \ \    1 \ + \ 2 \ \cos \left( \frac{2 \pi}{n} \right) \   = \    1 \ + \ \left[ \ \frac{\sin \left(   \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right] $$ $$\Rightarrow  \ \ 2 \ \sin \left(   \frac{2 \pi}{n} \right) \ \cos \left( \frac{2 \pi}{n} \right) \   = \ \sin \left(   \frac{3 \pi}{n} \right) \ \ \Rightarrow \ \ \sin \left(   \frac{4 \pi}{n} \right) \  = \ \sin \left(   \frac{3 \pi}{n} \right) \ \ , $$
applying the "double-angle formula" for sine at the end.
Since all of the angles discussed here have measure less than $ \ \pi \ $ , it is either the case that 
$$ \frac{4 \pi}{n}   \     =       \frac{3 \pi}{n} \ \ , $$
which would require $ \ \frac{  \pi}{n} \ = \ 0 \ $ , or that
$$ \frac{4 \pi}{n}   \     =    \ \pi \ -   \frac{3 \pi}{n} \ \ \Rightarrow \ \ \frac{7 \pi}{n}   \     =    \ \pi \ \ \Rightarrow \ \ n \ = \ 7 \ \ .  $$
