Shortcut for calculating pH of mixtures
Case 1: Mixing equal volumes of two solutions with a pH difference of 1.
Initially we have:
\begin{array}{|c|c|c|}
\hline 
&amp;[\ce{H+}]&amp; \mathrm{pH} &amp;n_\ce{H+}\\ \hline
a&amp; x &amp; \log x &amp; xV\\
b&amp;\frac{x}{10} &amp; -\log x + 1 &amp; \frac{xV}{10}\\
\hline
\end{array}
\begin{align}
\overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b}{2}=\frac{-\log x-\log x+1}{2}=-\log x+0.5
\end{align}
After complete mixing, we obtain:
$$
({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}=xV+\frac{xV}{10}=\frac{11xV}{10}\\
[\ce{H+}]_\mathrm{mix}=\frac{1.1xV}{2V}=\frac{11x}{20}\\
\mathrm{pH}_\mathrm{mix}=-\log \frac{11x}{20}=-\log x - \log\frac{11}{20}=\overline{\mathrm{pH}}-0.5+0.2596=\overline{\mathrm{pH}}-0.2404
$$
Case 2: Mixing equal volumes of three solutions with consecutive pH differences of 1.
Before mixing:
\begin{array}{|c|c|c|}
\hline 
&amp;[\ce{H+}]&amp; \mathrm{pH} &amp;n_\ce{H+}\\ \hline
a&amp; x &amp; \log x &amp; xV\\
b&amp;\frac{x}{10} &amp; -\log x + 1 &amp; \frac{xV}{10}\\
c&amp;\frac{x}{100} &amp; -\log x + 2 &amp; \frac{xV}{100}\\
\hline
\end{array}
\begin{align}
\overline{\mathrm{pH}}=\frac{\mathrm{pH}_a+\mathrm{pH}_b+\mathrm{pH}_c}{3}=\frac{-\log x-\log x+1-\log x+2}{3}=-\log x+1
\end{align}
After complete mixing, we obtain:
$$
({n_{\ce{H+}}})_\mathrm{mix}=({n_{\ce{H+}}})_{a}+({n_{\ce{H+}}})_{b}+({n_{\ce{H+}}})_{c}=\frac{111xV}{100}\\
[\ce{H+}]_\mathrm{mix}=\frac{1.11xV}{3V}=\frac{111x}{300}\\
\mathrm{pH}_\mathrm{mix}=-\log \frac{111x}{300}=-\log x - \log\frac{111}{300}=\overline{\mathrm{pH}}-1+0.4318=\overline{\mathrm{pH}}-0.5682
$$
(It seems you got the sign wrong on your last equation.)
Caveats:
Technically these calculations only work well if we assume that the activity of the solvated proton is the same as its concentration, which is a poor approximation below $\mathrm{pH}$ of 0 and above $\mathrm{pH}$ of 14.
You also get a slight deviation if you mix solutions that have $\mathrm{pH}$ too close to neutral, because the auto-dissociation of water affects the equilibrium values of $[\ce{H+}]$ and $[\ce{OH-}]$.
For example, if we mix two equal amounts of liquid, one being pure water($\mathrm{pH} = 7$) and the other being aqueous $\ce{HCl}$ solution with $[\ce{HCl}] = \pu{10^{-6}M}$. Considering the auto-dissociation constant for water as $K_\mathrm{w} = 10^{-14}$, the $\mathrm{pH}$ for the last solution is actually approximately $5.9957$ because water auto-dissociation already affects it slightly.
Neglecting that, we mix the liquids creating a solution with $[\ce{HCl}] = \pu{5\times 10^{-7} M}$. Directly using $[\ce{HCl}] = \pu{5\times 10^{-7}M} = [\ce{H+}]$ yields a $\mathrm{pH}$ of $6.3010$.
Using the trick, we get the pH for $[\ce{HCl}] = \pu{5.5\times 10^{-7}M}$ as $6.2596$. However, the real value for $[\ce{H+}]$, taking into account the shifting of the water auto-dissociation equilibrium, is $[\ce{H+}] = \pu{5.19258\times 10^{-7}M}$, such that the real $\mathrm{pH}$ is $6.28461$. A small deviation surely, but it's present.