Step-by-step Solution:

Given: The sequence \( a_1, a_2, a_3, \dots, a_n \) is an arithmetic progression (A.P.) with \( a_1 = 0 \). Let \( d \) be the common difference of the A.P. Then, \[ a_2 = a_1 + d = d \] \[ a_3 = a_1 + 2d = 2d \] \[ a_4 = 3d, \quad a_5 = 4d, \quad \dots, \quad a_n = (n-1)d \] We need to evaluate the given expression: \[ \frac{a_3}{a_2} + \frac{a_4}{a_3} + \dots + \frac{a_n}{a_{n-1}} - a_2 \left( \frac{1}{a_2} + \frac{1}{a_3} + \dots + \frac{1}{a_{n-2}} \right) \] Step 1: Evaluating the First Summation \[ \frac{a_3}{a_2} + \frac{a_4}{a_3} + \dots + \frac{a_n}{a_{n-1}} \] \[ = \frac{2d}{d} + \frac{3d}{2d} + \frac{4d}{3d} + \dots + \frac{(n-1)d}{(n-2)d} \] \[ = 2 + \frac{3}{2} + \frac{4}{3} + \dots + \frac{n-1}{n-2} \] Step 2: Evaluating the Second Summation \[ a_2 \left( \frac{1}{a_2} + \frac{1}{a_3} + \dots + \frac{1}{a_{n-2}} \right) \] \[ = d \left[ \frac{1}{d} + \frac{1}{2d} + \frac{1}{3d} + \dots + \frac{1}{(n-3)d} \right] \] \[ = \left[ 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n-3} \right] \] Step 3: Final Calculation \[ \left( (1+1) + \left( 1 + \frac{1}{2} \right) + \dots + \left( 1 + \frac{1}{n-2} \right) \right) \] \[ - \left( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n-3} \right) \] \[ = 1 + 1 + 1 + \dots \text{ (up to } (n-2) \text{ terms)} \] \[ - \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{(n-3)} - \frac{1}{(n-2)} \right) \] \[ = (n-2) + \frac{1}{n-2} \] Thus, the final answer is: \[ \boxed{(n-2) + \frac{1}{n-2}} \]