A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time $$t_1$$. If he remains stationary on a moving escalator then the escalator takes him up in time $$t_2$$. The time taken by him to walk up on the moving escalator will be:

Let the length of the escalator be $$L$$. The boy's walking speed is $$v_b = L/t_1$$ and the escalator's speed is $$v_e = L/t_2$$.

When the boy walks on the moving escalator, the effective speed is $$v_b + v_e = \frac{L}{t_1} + \frac{L}{t_2} = L\cdot\frac{t_1 + t_2}{t_1 t_2}$$.

The time taken is $$t = \frac{L}{v_b + v_e} = \frac{L \cdot t_1 t_2}{L(t_1 + t_2)} = \frac{t_1 t_2}{t_1 + t_2}$$.

Therefore the time taken by him to walk up on the moving escalator is $$\dfrac{t_1 t_2}{t_1 + t_2}$$.