Why does the half time decrease in a 0th order reaction?

A --> B


rate = k


Half time = [A]0/2k 


From the half time equation, since there are no variables, i don't see how the half time of a 0th order reaction can change? It's dependent on the initial concentration of A, which is constant, so how can the half time be decreasing as shown in the graph below.


[![enter image description here](https://i.stack.imgur.com/TNu1G.png)](https://i.stack.imgur.com/TNu1G.png)


The only explanation I can imagine is that A0 is not the initial concentration at the start of the reaction, but the concentration at each the start of each half time.




For the first half life(say,$T^1\_{\frac{1}{2}}$ )and the second half life(say, $T^2\_{\frac{1}{2}}$) the initial concentrations of the reactant to be used for calculating are different.   
For, the $0$th order reaction $\ce{A -> B}$, the integrated rate equation will be, $$\ce{[A]\_t = [A]\_0 - kt}$$ where $k$ is the rate constant.  
So, at the first half life($t$ = $T^1\_{\frac{1}{2}}$), $\ce{[A]\_t = \frac{[A]\_0}{2}}$, So, we have $T^1\_{\frac{1}{2}}$= $\frac{[A]\_0}{2k}$   
But for the second half life ($t$ = $T^2\_{\frac{1}{2}}$), $\ce{[A]\_t^, = \frac{[A]\_0}{4}}$ and $\ce{[A]\_0^' = \frac{[A]\_0}{2}}$, Putting these values in the equation, we have $T^2\_{\frac{1}{2}}$ = $\frac{[A]\_0}{4k}$  
 Thus we can see that for any $nth$ half life in a $0$ th order reaction, the $[A]\_0$ which we are habituated to use in $1st$ half life is not the initial concentration(i.e. at $t=0$) but for the nth half life, $[A]\_0$ is the concentration of the reactant after completion of $n-1$ half lives i.e $\frac{[A]\_0}{2^{n-1}}$, and thus using this fact a general formula for nth half life (i.e. $T^n\_{\frac{1}{2}}$) in ) $0$ th order reaction will be $\frac{[A]\_0}{2^nk}$.


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