Harmonic mean (H.M.)

If $x$ and $y$ are two numbers, their Harmonic Mean is $H=\dfrac{2xy}{x+y}$. This is the value that makes $x, H, y$ an H.P., since $\dfrac1x,\dfrac1H,\dfrac1y$ must be in A.P.: $\dfrac1H=\dfrac{\frac1x+\frac1y}2=\dfrac{x+y}{2xy} \Rightarrow H=\dfrac{2xy}{x+y}$.

These extend to $n$ positive numbers as $A=\dfrac{x_1+x_2+\cdots+x_n}n$, $G=\left(x_1x_2\cdots x_n\right)^{1/n}$, $H=\dfrac n{\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_n}}$. Also, if $x=y$ then $A=G=H$.

Theorem: if $A,G,H$ are the A.M., G.M., H.M. of two positive numbers $x$ and $y$, then (i) $G^2=AH$, and (ii) $A\ge G\ge H$.

Proof: $A=\dfrac{x+y}2, G=\sqrt{xy}, H=\dfrac{2xy}{x+y}$. (i) $AH=\dfrac{x+y}2\cdot\dfrac{2xy}{x+y}=xy=G^2$. (ii) Consider $A-G=\dfrac{x+y}2-\sqrt{xy}=\dfrac12\left(x+y-2\sqrt{xy}\right)=\dfrac12\left(\sqrt x-\sqrt y\right)^2\ge0$ since a square is always non-negative; hence $A\ge G$ ...(I), so $\dfrac AG\ge1$ ...(II). Since $G^2=AH$, $\dfrac GH=\dfrac AG\ge1$ (from II), so $G\ge H$ ...(III). From (I) and (III), $A\ge G\ge H$.

Worked Example 1: find A.M., G.M., H.M. of 4 and 16. $A=\dfrac{4+16}2=10$. $G=\sqrt{4\times16}=\sqrt{64}=8$. $H=\dfrac{2\times4\times16}{4+16}=\dfrac{128}{20}=\dfrac{32}5$.

Worked Example 2: insert 4 terms between 2 and 22 so the new sequence is an A.P. Let $A_1,A_2,A_3,A_4$ be the terms, so $2,A_1,A_2,A_3,A_4,22$ is an A.P. with $a=2,\ t_6=22,\ n=6$: $22=2+5d \Rightarrow d=4$. $A_1=6, A_2=10, A_3=14, A_4=18$.

Worked Example 3: insert two numbers between $\dfrac29$ and $\dfrac1{12}$ so the resulting sequence is an H.P. Let the numbers be $\dfrac1{H_1},\dfrac1{H_2}$, so $\dfrac92,H_1,H_2,12$ is an A.P. with $a=\dfrac92, t_4=12$: $\dfrac92+3d=12 \Rightarrow d=\dfrac52$. $H_1=\dfrac92+\dfrac52=7$, $H_2=\dfrac92+2\left(\dfrac52\right)=\dfrac{19}2$. The inserted numbers are $\dfrac17$ and $\dfrac2{19}$.

Worked Example 4: insert two numbers between 1 and 27 so the resulting sequence is a G.P. $1,G_1,G_2,27$ G.P. with $a=1, t_4=27$: $r^3=27 \Rightarrow r=3$. $G_1=3, G_2=9$. …