If $a_1,a_2,\ldots,a_n$ are in A.P. and $a_1=0$ then the value of $\left(\dfrac{a_3}{a_2}+\dfrac{a_4}{a_3}+\cdots+\dfrac{a_n}{a_{n-1}}\right)-a_2\left(\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_{n-2}}\right)$ is equal to