If the body covers equal displacements in successive intervals of time  $t_1,t_2$ and $t_3$ in a straight path then shown that $\frac{1}{t_1}-\frac{1}{t_2}+\frac{1}{t_3}=\frac{k}{t_1+t_2+t_3}$. Find k

Let velocities at A, B, C and D be $V_A,V_B,V_C$ and $V_D$ respectively

For AB $l=\left(\frac{V_A+V_B}{2}\right)t_1$  ...(1)
For BC $l=\left(\frac{V_B+V_C}{2}\right)t_2$  …(2) 

For CD $l=\left(\frac{V_C+V_D}{2}\right)t_3$  …(3)      
     From (1), (2) and (3), we have $\frac{1}{{\mathrm{t}}_1}-\frac{1}{{\mathrm{t}}_2}+\frac{1}{{\mathrm{t}}_3}=\frac{1}{2l}\left({\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{C}}+{\mathrm{V}}_{\mathrm{C}}+{\mathrm{V}}_{\mathrm{D}}\right)\Rightarrow \frac{1}{{\mathrm{t}}_1}-\frac{1}{{\mathrm{t}}_2}+\frac{1}{{\mathrm{t}}_3}=\frac{1}{2l}\left({\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{D}}\right)$.....(4)
$\text{ AD }3l=\left(\frac{{\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{D}}}{2}\right)\left({\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3\right)\Rightarrow \frac{1}{2l}\left({\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{D}}\right)=\frac{3}{{\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3}\dots$(5)
So, from (4) and (5), we get
$\Rightarrow k=3$