Dimensional consistency of an equation We know that if an equation has to be physically correct then it must be dimensionally consistent i.e. If an equation is not dimensionally correct then it can never be physically correct. Now in the equation of distance travelled by a particle in nth second, we see distance at LHS and the sum of $u$ and $a/2(2n-1)$ on the RHS. Now how can this equation be dimensionally consistent as it doesn't seem to obey principle of homogeneity (the dimensions of distance and velocity $u$ are not the same). And if this equation is dimensionally incorrect so it shouldn't be correct physically. 
 A: You're correct: whatever your teacher has shown you is dimensionally incorrect, and comes from the assumption of breaking the world into one-second "time steps." 
The correct equation for the position of a particle undergoing constant acceleration $a$ with position $x_0$ and velocity $v_0$ at time $t=0$ is $$x(t) = x_0 + v_0~t + \frac12~a~t^2.$$Your formula appears to come from substituting $u = v_0,$ with $x_n = x(t_n)$ and $t_n = n~\delta t$ for some time step $\delta t$ (probably $\delta t = 1\text{ s}$). Then the change in position over one time step is: $$x_n - x_{n-1} =\left(x_0 + u~n~\delta t + \frac12~a~n^2~\delta t^2\right) - \left(x_0 + u~(n - 1)~\delta t + \frac12~a~(n^2 - 2n + 1)~\delta t^2\right),$$and the leading terms all cancel to give $$x_n - x_{n-1} = v_0~\delta t + \frac12~a~(2n - 1)~\delta t^2.$$
First, a caveat: this is not necessarily "the distance traveled" during that timestep; even $|x_n - x_{n-1}|,$ which properly accounts for the possible minus sign(!), still fails to truly be this distance during any timestep where $t_{n-1} < (-u/a) < t_n,$ when the velocity changes sign. Only if this happens to be on a timestep boundary or outside the time bounds that you are interested in, can we use $|x_n - x_{n-1}|$ as the true distance covered in the $n^\text{th}$ time step; and even then, only if $x_n - x_{n-1}|$ is positive will your expression be totally correct.
The reason that you see something "dimensionally inconsistent" is because the units from the $\delta t$ have been absorbed into the corresponding other constants -- in other words the equation that you're looking at has redefined $a~\delta t^2\mapsto a$ and $u~\delta t\mapsto u.$ It has probably done this because $\delta t = 1\text{ s}$ and therefore if you work entirely in SI units then the numbers that you punch into the calculator are the same either way.
This is a tedious problem which also, after a fashion, affects very high-end physics work: it is very common to see authors in high-end papers say "using units where $\hbar = c = 1,$ this equation is...". This sloppiness about the dimensionality of things makes it really easy to write theory papers but very hard for experimentalists to follow up and say "okay, here's what we are trying to measure, here's what our results should look like," etc. So please regard this as good training for a career in theoretical physics, not a sloppy slip-up by your teacher!
A: The equation is perfectly correct . Lets see its derivation . 
The equation for the position of a particle undergoing constant acceleration $a$ with position $x_0$ and velocity $v_0$ at time instant $t$ is $$x(t) = x_0 + v_0~t + \frac12~a~t^2.$$The formula comes from substracting the position at time (t-1) seconds. Then : $$x_t - x_{t-1} =\left(x_0 + ut + \frac12~a~t^2\right) - \left(x_0 + u( t-1 s) + \frac12~a~(t^2 - 2t + 1)\right),$$and the terms cancel to give $$x_t - x_{t-1} = v_0( 1 s ) + \frac12~a~(2t - 1  s)~( 1 s).$$
As you can see we generally skip writing the units. So dont do it. Additionally the equation is now dimensionally correct.
