$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.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?

This problem can be trivialized by Ptolemy's theorem.
First we take the l.c.m and simplify both sides of the given equation.
Let $A_1A_2=a$, $A_1A_3=b$, $A_1A_4=c$. Then we have 
$$\frac{1}{a}=\frac{1}{b}+\frac{1}{c} \qquad\to\qquad b c = a b + a c \tag{1}$$
Now, note that regular polygons can always be inscribed in a circle.
Take the quadrilateral $A_1A_3A_4A_5$, which is a cyclic quadrilateral.
Applying Ptolemy's theorem we get,
$$A_1A_3\cdot A_4A_5 + A_3A_4\cdot A_1A_5 = A_1A_4\cdot A_3A_5 \tag{2}$$
Now see that $A_1A_3=b$, $A_4A_5=a$, $A_3A_4=a$, $A_1A_4=c$, $A_3A_5=b$,
so in $(2)$ we have,
$$b a + a\cdot A_1A_5 = c b \tag{3}$$
Comparing equation $(1)$ with $(3)$, we see that $A_1A_5 = c = A_1A_4$.
These are two diagonals of the regular polygon sharing a common vertex, so this diagonal must be a diagonal that bisects the area of the polygon. So on one side of the diagonal $A_1A_4$ there is $A_2$, $A_3$; and on the other side of the diagonal $A_1A_5$ there must be $A_6$, $A_7$. So there are 7 vertices of the polygon. 

Hence $n=7$, leading to a heptagon.

