A train takes t sec to perform a journey, if travel for t/n sec with uniform acceleration then for $\left(\frac{\mathrm{n}-3}{\mathrm{n}}\right)\mathrm{t}$ sec with uniform speed v and finally it comes to rest with uniform retardation. Then average speed of train is 

Total time = tsec;  First interval = t/n ;

second interval = $\left(\frac{\mathrm{n}-3}{\mathrm{n}}\right)\mathrm{t}$

third interval = 2t/n
$\begin{array}{l}{\mathrm{s}}_1=\frac{1}{2}\mathrm{a}{\left(\frac{\mathrm{t}}{\mathrm{n}}\right)}^2;{\mathrm{s}}_2=\mathrm{v}\left(\frac{\mathrm{n}-3}{\mathrm{n}}\right)\mathrm{t}\\ {\mathrm{s}}_3=\mathrm{v}\cdot \left(\frac{2\mathrm{t}}{\mathrm{n}}\right)-\frac{1}{2}\mathrm{b}\cdot \frac{4{\mathrm{t}}^2}{{\mathrm{n}}^2};{\mathrm{v}}_{\text{average }}=\frac{{\mathrm{s}}_1+{\mathrm{s}}_2+{\mathrm{s}}_3}{\mathrm{t}}\\ {\mathrm{v}}_{\text{average }}=\frac{\frac{1}{2}\mathrm{a}{\left(\frac{\mathrm{t}}{\mathrm{n}}\right)}^2+\mathrm{v}\left(\frac{\mathrm{n}-3}{\mathrm{n}}\right)\mathrm{t}+\mathrm{v}\cdot \left(\frac{2\mathrm{t}}{\mathrm{n}}\right)-\frac{1}{2}\cdot \mathrm{b}\frac{4{\mathrm{t}}^2}{{\mathrm{n}}^2}}{\mathrm{t}}\end{array}$

from v = u + at

$\mathrm{v}=\mathrm{o}+\mathrm{a}\left(\frac{\mathrm{t}}{\mathrm{n}}\right);\mathrm{a}=\frac{\mathrm{nv}}{\mathrm{t}}$ (first interval)
$\mathrm{o}=\mathrm{v}-\mathrm{b}\left(\frac{2\mathrm{t}}{\mathrm{n}}\right);\mathrm{b}=\frac{\mathrm{nv}}{2\mathrm{t}}$ (third interval)

substitute a & b in above equation

$\begin{array}{l}{\mathrm{v}}_{\text{average }}=\frac{\mathrm{v}}{\mathrm{n}}\left[\frac{3+2\mathrm{n}-6}{2}\right]=\frac{\mathrm{v}}{\mathrm{n}}\left[\frac{2\mathrm{n}-3}{2}\right]\\ =\frac{\mathrm{v}}{2\mathrm{n}}(2\mathrm{n}-3)\end{array}$