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Consider two hypothetical parcels at the same temperature $300\mathrm{K}$, kept separated with negligible partition mass and excluded container volume. Parcel A has mass $2\mathrm{kg}$ and volume $0.002{\mathrm{m}}^3$; parcel B has mass $3\mathrm{kg}$ and volume $0.006{\mathrm{m}}^3$. They may be different materials. Their individual mass densities and the composite mass density areThe composite mass density is total material mass divided by total occupied volume; it need not be the local mass density anywhere. Equal temperature does not make these parcels replicas. In particular, the unweighted average $750\mathrm{kg}{\mathrm{m}}^{-3}$ is wrong. The volume fractions are $1/4$ and $3/4$, givingWith mass fractions $2/5$ and $3/5$, the equivalent rule is a weighted harmonic mean:This follows directly by writing $V_i=m_i/{\rho }_i$ and dividing total volume by total mass; it is also the mass-weighted average of specific volumes. An unweighted average of mass densities is valid when the parcel volume values are equal, not merely when their temperatures are equal.

If a partition is removed and different substances mix, the final volume need not equal the sum of initial volume values. A new temperature, thermodynamic phase or composition may also change the local mass density. Use the actual final volume in $\rho =m/V$; the pre-mixing weighted identity cannot supply it. Even unchanged temperature alone does not guarantee unchanged volume.

The harmonic mean of a nonempty finite list of positive real numbers is the multiplicative inverse of the arithmetic mean of their inverses:For two inputs, $H=2x_1{x}_2/(x_1+x_2)$. Replacing each input by $H$ preserves the sum of inverses, making it useful when a fixed amount divided by a rate determines a cost or duration. This positive-input definition avoids division by zero; negative inputs need not retain the usual averaging or ordering properties. ^[ref].

Travel two legs of equal length $d$ at positive speeds $v_1,v_2$, with no stops. Their travel times are $d/v_1$ and $d/v_2$, so the average speed isFor two $60\mathrm{km}$ legs at $30$ and $60km/h$, the times are two hours and one hour. The average speed is $120/3=40km/h$. The same speeds maintained for one hour each instead cover $90\mathrm{km}$ in two hours, giving the arithmetic mean $45km/h$. A slower equal-distance leg occupies more time, so the data being averaged must include what is held equal. This is total distance traveled divided by elapsed time; a return journey can have zero average velocity while keeping a positive average speed. ^[ref].

The arithmetic-geometric mean inequality applied to the inverses gives $H\le G$; together with $G\le A$ it yields the comparison in geometric mean. For two inputs the arithmetic–harmonic gap has the exact formFor $30$ and $60$ the gap is five. With one positive input held fixed and another approaching zero, the harmonic mean approaches zero too: a very slow fixed-distance leg dominates the travel time. Unequal leg lengths instead require a weighted harmonic mean.

An isotopic abundance is an atom fraction within a specified population of one chemical element.^[ref] If $n_i$ counts its chemical atoms of isotope $i$, useThe denominator includes every isotope of that chemical element included in the stated population, not every chemical atom of every chemical element in the sample. Thus an oxygen atom fraction among all chemical elements is a different quantity.

For an illustrative neutral neon population, take the displayed central isotopic abundances $f_{20}=0.9048$, $f_{21}=0.0027$, and $f_{22}=0.0925$ from the reference table, so $0.9048+0.0027+0.0925=1$. Its mean atomic mass is the weighted arithmetic meanThese are rounded reference inputs with uncertainties, not exact universal isotopic abundances; the long decimal shows their arithmetic result, not measurement precision.^[ref]

The population convention matters. If instead $w_i$ is the fraction of total neutral-atom mass supplied by isotope $i$, then $w_i=f_i{m}_i/\bar{m}$ andThis is a weighted harmonic mean in the mass fractions. Using ${\sum }_i{w}_i{m}_i$ with those mass fractions would answer a different question. The conversion follows directly by counting chemical atoms and adding their masses.

For positive real numbers $x_i$ and nonnegative weights with ${\sum }_i{w}_i=1$, the weighted geometric mean isIt takes a weighted mean of logarithms. A zero weight excludes that input; equal weights give the ordinary geometric mean. If $x_i$ occurs $m_i$ times in a list of total length $M$, its correct weight is $w_i=m_i/M$. For the list $(12,3,3,3)$,and replacing all four entries by $G_w$ preserves the product $324$.

There is also an optimization interpretation. A logarithmic transformation turns multiplicative discrepancies into additive ones. For a positive representative $c$, the weighted mean minimizes squared deviations identity givesThus $c=G_w$ uniquely minimizes weighted quadratic loss in logarithmic coordinates. The weighted arithmetic mean minimizes squared discrepancies in the original coordinates. These two objectives explain why different choices of what constitutes an error select different summaries.

The strictly concave function $\log x$ and the reversed Jensen's inequality give $G_w\le A_w$, where $A_w={\sum }_i{w}_i{x}_i$ is the weighted arithmetic mean. Apply the same reasoning to the multiplicative inverses to get $H_w\le G_w$ for the weighted harmonic mean. Thus $H_w\le G_w\le A_w$; equality holds exactly when all positive-weight inputs agree. For this list, $H_w=48/13$, $G_w=3\sqrt{2}$, and $A_w=21/4$. The weights specify which inputs are represented and must stay the same across the comparison.