Exercises — Harmonic progression — definition, HM

What we do: flip every term, because HP is defined through reciprocals — we can never judge it directly. Reciprocals: $\frac{1}{12},\frac{1}{6},\frac{1}{4},\frac{1}{3}$. Common denominator $12$: $\frac{1}{12},\frac{2}{12},\frac{3}{12},\frac{4}{12}$. Check the gaps: each step adds $\frac{1}{12}$. Constant! ✅ (This is exactly the "flip up to the even ruler" move from the opening figure.) So the reciprocals are an AP with $d=\frac{1}{12}$ ⇒ the originals are in HP. Yes.

$0$ can never appear. You would have to take the reciprocal $\frac{1}{0}$, which is undefined. $-5$ and $\frac{1}{2}$ are both fine — negatives and fractions have perfectly good reciprocals.

Flip: reciprocals are $\frac{1}{3},\frac{2}{3},1,\dots$ i.e. $\frac{1}{3},\frac{2}{3},\frac{3}{3}$ — an AP with $A=\frac{1}{3}$, $D=\frac{1}{3}$. AP $6^{\text{th}}$ term (why: this is the reciprocal of what we want): $A+(6-1)D=\frac{1}{3}+5\cdot \frac{1}{3}=\frac{6}{3}=2.$ Flip back: $a_6=\frac{1}{2}$.

Flip: the reciprocals form an AP with $\frac{1}{a_2}=5$ and $\frac{1}{a_5}=11$. Two AP terms $\Rightarrow$ find $D$. From term 2 to term 5 is $3$ steps: $11=5+3D\Rightarrow D=2.$ Step back from term 2 to term 1. Recall the AP nth-term rule $\frac{1}{a_n}=A+(n-1)D$. Writing it for $n=2$ and $n=1$: $\frac{1}{a_2}=A+D,\frac{1}{a_1}=A.$ Subtracting these gives $\frac{1}{a_2}-\frac{1}{a_1}=D$, so $\frac{1}{a_1}=\frac{1}{a_2}-D$ — i.e. moving down one index means subtracting one $D$ (the "travel one even step back down the ruler" move from the opening figure). Numerically: $\frac{1}{a_1}=5-D=5-2=3.$ Flip back: $a_1=\frac{1}{3}$.

Use TABS (the HM formula $\frac{2ab}{a+b}$ = Two·A·B·over·Sum, introduced above): $H=\frac{2ab}{a+b}=\frac{2\cdot 4\cdot 12}{4+12}=\frac{96}{16}=6.$ Sanity flip-check: reciprocals $\frac{1}{4},\frac{1}{6},\frac{1}{12}$ = $0.25,0.1\overline{6},0.08\overline{3}$; both gaps $=-0.08\overline{3}$. ✅ AP.

What is $H$? The HM, so $\frac{1}{a},\frac{1}{H},\frac{1}{b}$ form an AP (equal gaps). Equal-gap equation: $\frac{1}{H}-\frac{1}{a}=\frac{1}{b}-\frac{1}{H}$, so $\frac{1}{b}=\frac{2}{H}-\frac{1}{a}$. $\frac{1}{b}=\frac{2}{4}-\frac{1}{3}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\Rightarrow b=6.$ Check with TABS ($\frac{2ab}{a+b}$): $\frac{2\cdot 3\cdot 6}{3+6}=\frac{36}{9}=4=H.$ ✅

Read the figure first. All three coloured legs on the top track have the same length — that is the "equal distances" condition. The bars hanging below have different lengths: the slow $30$ km/h leg has the longest time bar, the fast $60$ km/h leg the shortest. Total trip time is the three bars stacked; it is these bar-lengths (the times $d/\text{speed}$) that add, which is exactly why the harmonic mean rules here. Why HM? The legs are equal in distance, not equal in time. Call each leg length $d$ (same for all three). For each leg $\text{time}=\frac{d}{\text{speed}}$, so it is the reciprocals of speed that add — exactly the harmonic setup drawn above. Start from the definition of average speed (total distance over total time — this always holds): $\bar{s}=\frac{\text{total distance}}{\text{total time}}=\frac{d+d+d}{\frac{d}{30}+\frac{d}{40}+\frac{d}{60}}=\frac{3d}{\frac{d}{30}+\frac{d}{40}+\frac{d}{60}}.$ Now $d$ cancels. Factor $d$ out of every term in the denominator: $\frac{d}{30}+\frac{d}{40}+\frac{d}{60}=d\left(\frac{1}{30}+\frac{1}{40}+\frac{1}{60}\right)$. Then the $3d$ on top and the $d$ in the bracket cancel a common factor $d$: $\bar{s}=\frac{3d}{d\left(\frac{1}{30}+\frac{1}{40}+\frac{1}{60}\right)}=\frac{3}{\frac{1}{30}+\frac{1}{40}+\frac{1}{60}}.$ Why $d$ vanishes: the leg length appears once on top and once in each bottom term (each bar in the figure scales with $d$), so it never affects the average — that's why we never needed a numerical distance. Add the reciprocals (common denominator $120$): $\frac{4}{120}+\frac{3}{120}+\frac{2}{120}=\frac{9}{120}=\frac{3}{40}$. Flip and multiply by $3$: $\bar{s}=\frac{3}{3/40}=40$ km/h.

AM $=\frac{a+b}{2}=\frac{20}{2}=10.$ GM $=\sqrt{ab}=\sqrt{36}=6.$ HM $=\frac{2ab}{a+b}=\frac{2\cdot 2\cdot 18}{20}=\frac{72}{20}=\mathrm{3.6.}$ Ordering: $10\ge 6\ge 3.6$. ✅ Product identity: $\text{AM}\cdot \text{HM}=10\cdot 3.6=36=6^2={\text{GM}}^2.$ ✅

Flip: reciprocals form an AP with $\frac{1}{a_3}=\frac{1}{7}$ and $\frac{1}{a_7}=\frac{1}{3}$. Find $D$ (from term 3 to term 7 is $4$ steps): $\frac{1}{3}=\frac{1}{7}+4D\Rightarrow 4D=\frac{1}{3}-\frac{1}{7}=\frac{7-3}{21}=\frac{4}{21}\Rightarrow D=\frac{1}{21}.$ First reciprocal (step back 2 from term 3): $\frac{1}{a_1}=\frac{1}{7}-2D=\frac{3}{21}-\frac{2}{21}=\frac{1}{21}$, so $a_1=21$. $10^{\text{th}}$ reciprocal: $\frac{1}{a_{10}}=\frac{1}{21}+9\cdot \frac{1}{21}=\frac{10}{21}$, so $a_{10}=\frac{21}{10}=2.1$.

Set up on the reciprocals (call them $x-d,x,x+d$, an AP — that's what HP guarantees). Sum of reciprocals $=3x=\frac{3}{4}\Rightarrow x=\frac{1}{4}$. So the middle term is $a_2=\frac{1}{x}=4$. We still need $d$. Use the sum of the originals $=11$: $\frac{1}{x-d}+\frac{1}{x}+\frac{1}{x+d}=11.$ With $x=\frac{1}{4}$: $\frac{1}{1/4-d}+4+\frac{1}{1/4+d}=11\Rightarrow \frac{1}{1/4-d}+\frac{1}{1/4+d}=7.$ Combine the two fractions. Put them over the common denominator $(1/4-d)(1/4+d)$: $\frac{1}{1/4-d}+\frac{1}{1/4+d}=\frac{(1/4+d)}{(1/4-d)(1/4+d)}+\frac{(1/4-d)}{(1/4-d)(1/4+d)}=\frac{(1/4+d)+(1/4-d)}{(1/4-d)(1/4+d)}.$ The numerator adds to $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ (the $+d$ and $-d$ cancel); the denominator is a difference of squares $(1/4{)}^2-d^2=\frac{1}{16}-d^2$. So the equation becomes $\frac{1/2}{1/16-d^2}=7.$ Hence $1/16-d^2=\frac{1/2}{7}=\frac{1}{14}$, giving $d^2=\frac{1}{16}-\frac{1}{14}=\frac{7-8}{112}=-\frac{1}{112}.$ $d^2<0$ — impossible! No such real HP exists with these two conditions; they contradict each other. What this teaches: the sum and reciprocal-sum can't be chosen independently; requiring both forces $d^2<0$.

Restate with formulas: show $\frac{2ab}{a+b}<\sqrt{ab}$. Why start from AM $\ge$ GM? It's the cleanest known inequality, and $\text{HM}=\frac{ab}{\text{AM}}$ ties them together. Step 1 — AM $>$ GM for distinct positives: $\frac{a+b}{2}>\sqrt{ab}$ (strict because $a\ne b$; comes from $(\sqrt{a}-\sqrt{b}{)}^2>0$). Step 2 — divide the fixed positive number $ab$ by each side. Here is the careful justification: if $0<P<Q$ are two positive quantities and $M>0$ is a fixed positive number, then $\frac{M}{P}>\frac{M}{Q}$. Why? $\frac{M}{P}-\frac{M}{Q}=\frac{M(Q-P)}{PQ}$, and every factor $M,(Q-P),P,Q$ is positive, so the difference is positive. Dividing a fixed number by a larger denominator gives a smaller result — no inequality is reversed, because we are not dividing both sides of the inequality by the same thing; we are forming two new fractions with the same fixed numerator $M=ab$ and different denominators. Apply this with $M=ab>0$, $P=\sqrt{ab}$ (the smaller side, GM), $Q=\frac{a+b}{2}$ (the larger side, AM): $\frac{ab}{(a+b)/2}<\frac{ab}{\sqrt{ab}}.$ Step 3 — simplify both sides: left $=\frac{2ab}{a+b}=\text{HM}$; right $=\frac{ab}{\sqrt{ab}}=\sqrt{ab}=\text{GM}$. Therefore $\text{HM}<\text{GM}$. ∎ (Equality only when $a=b$.)

Flip: the reciprocals form an AP. Its $n^{\text{th}}$ term is $t_n=A+(n-1)D$, and $t_n=\frac{1}{a_n}$. Translate the givens: $a_p=q$ means $t_p=\frac{1}{q}$; $a_q=p$ means $t_q=\frac{1}{p}$. $A+(p-1)D=\frac{1}{q},A+(q-1)D=\frac{1}{p}.$ Subtract (isolates $D$): $(p-q)D=\frac{1}{q}-\frac{1}{p}=\frac{p-q}{pq}$, so $D=\frac{1}{pq}$. Back-substitute for $A$: $A=\frac{1}{q}-(p-1)\frac{1}{pq}=\frac{p}{pq}-\frac{p-1}{pq}=\frac{1}{pq}$. The $(pq{)}^{\text{th}}$ reciprocal: $t_{pq}=A+(pq-1)D=\frac{1}{pq}+(pq-1)\frac{1}{pq}=\frac{pq}{pq}=1.$ Flip back: $a_{pq}=\frac{1}{1}=1$. A clean, memorable result: the $(pq{)}^{\text{th}}$ term is $\boxed{1}$.

Test for HP ::: Flip every term; check the reciprocals have a constant difference. $6^{\text{th}}$ term of $3,\frac{3}{2},1,\dots$ ::: $\frac{1}{2}$. HM of $4$ and $12$ ::: $6$. Average speed over 3 equal legs at $30,40,60$ ::: $40$ km/h (HM, not $43.\overline{3}$). $(pq{)}^{\text{th}}$ term when $a_p=q,a_q=p$ ::: $1$.