Harmonic progressions and the harmonic mean

The third kind of pro­gres­sion is defined through the first. Take an A.P. such as $2, 3, 4, 5$ and turn every term upside down: $\displaystyle \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$. The new sequence is called a har­monic pro­gres­sion. The name comes from music: strings whose lengths are in har­monic pro­gres­sion sound notes that are in har­mony. This les­son defines har­monic pro­gres­sions and the har­monic mean, and ends by com­par­ing all three means.

Def­i­n­i­tion#

A sequence is said to be in har­monic pro­gres­sion (H.P.) if the rec­i­p­ro­cals of its terms form an A.P.

So $\displaystyle \frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots$ is an H.P., because $2, 5, 8, 11, \ldots$ is an A.P. And $6, 4, 3$ is an H.P., because $\displaystyle \frac{1}{6}, \frac{1}{4}, \frac{1}{3}$ is an A.P. with com­mon dif­fer­ence $\displaystyle \frac{1}{12}$. No term of an H.P. can be 0, since 0 has no rec­i­p­ro­cal.

The gen­eral term#

If the rec­i­p­ro­cals form the A.P. $a, a + d, a + 2d, \ldots$, then the $n$th term of the H.P. is

$$\displaystyle t_n = \frac{1}{a + (n - 1)d}$$

There is no sim­ple for­mula for the sum of the terms of an H.P. Ques­tions on H.P.s are solved by mov­ing to the rec­i­p­ro­cals, work­ing with the A.P., and mov­ing back at the end.

Exam­ple 1#

Find the 10th term of the H.P. $\displaystyle 2, \frac{4}{3}, 1, \frac{4}{5}, \ldots$

The rec­i­p­ro­cals are $\displaystyle \frac{1}{2}, \frac{3}{4}, 1, \frac{5}{4}, \ldots$, an A.P. with first term $\displaystyle \frac{1}{2}$ and com­mon dif­fer­ence $\displaystyle \frac{1}{4}$. Its 10th term is $\displaystyle \frac{1}{2} + 9 \times \frac{1}{4} = \frac{11}{4}$. So the 10th term of the H.P. is $\displaystyle \frac{4}{11}$.

Exam­ple 2#

The 3rd term of an H.P. is $\displaystyle \frac{1}{7}$ and the 7th term is $\displaystyle \frac{1}{15}$. Find the 12th term.

In the A.P. of rec­i­p­ro­cals, $t_3 = 7$ and $t_7 = 15$. So $4d = 8$, $d = 2$, and the first term is $7 - 2 \times 2 = 3$. The 12th term of the A.P. is $3 + 11 \times 2 = 25$, so the 12th term of the H.P. is $\displaystyle \frac{1}{25}$.

Prop­erty 1: three num­bers in H.P.#

$a$, $b$, $c$ are in H.P. exactly when $\displaystyle \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P., that is when $\displaystyle \frac{2}{b} = \frac{1}{a} + \frac{1}{c} = \frac{a + c}{ac}$. Turn­ing this over:

$$\displaystyle a, b, c \text{ are in H.P.} \iff b = \frac{2ac}{a + c}$$

Exam­ple 3#

If $3$, $b$, $6$ are in H.P., find $b$.

$\displaystyle b = \frac{2 \times 3 \times 6}{3 + 6} = \frac{36}{9} = 4$. Check: $\displaystyle \frac{1}{3}, \frac{1}{4}, \frac{1}{6}$ have dif­fer­ences $\displaystyle -\frac{1}{12}$ and $\displaystyle -\frac{1}{12}$.

Prop­erty 2: a test by ratios#

$$\displaystyle a, b, c \text{ are in H.P.} \iff \frac{a - b}{b - c} = \frac{a}{c}$$

Why. Start from $\displaystyle \frac{1}{a} - \frac{1}{b} = \frac{1}{b} - \frac{1}{c}$, that is $\displaystyle \frac{b - a}{ab} = \frac{c - b}{bc}$. Mul­ti­ply both sides by $b$: $\displaystyle \frac{b - a}{a} = \frac{c - b}{c}$. Rear­rang­ing gives $\displaystyle \frac{a - b}{b - c} = \frac{a}{c}$.

Com­pare the sim­i­lar tests for the other pro­gres­sions: for an A.P. the ratio $\displaystyle \frac{a - b}{b - c}$ equals $\displaystyle \frac{a}{a} = 1$, and for a G.P. it equals $\displaystyle \frac{a}{b}$. For the H.P. $3, 4, 6$: $\displaystyle \frac{3 - 4}{4 - 6} = \frac{1}{2}$ and $\displaystyle \frac{3}{6} = \frac{1}{2}$.

Har­monic mean#

If $a$, $H$, $b$ are in H.P., then $H$ is called the har­monic mean (H.M.) of $a$ and $b$. By Prop­erty 1,

$$\displaystyle H = \frac{2ab}{a + b}$$

For exam­ple, the H.M. of 4 and 12 is $\displaystyle \frac{2 \times 4 \times 12}{16} = 6$.

The har­monic mean is the right aver­age for rates. If Ravi dri­ves from Vijayawada to Gun­tur at 30 km/h and returns along the same road at 60 km/h, his aver­age speed for the whole trip is not 45 km/h but the H.M. of 30 and 60: $\displaystyle \frac{2 \times 30 \times 60}{90} = 40$ km/h. (Check with a dis­tance of 30 km: the trip takes 1 hour out and half an hour back, so 60 km in 1.5 hours, which is 40 km/h.)

Insert­ing har­monic means#

To insert $n$ har­monic means between $a$ and $b$, insert $n$ arith­metic means between $\displaystyle \frac{1}{a}$ and $\displaystyle \frac{1}{b}$, then take rec­i­p­ro­cals.

Exam­ple 4#

Insert two har­monic means between $\displaystyle \frac{1}{2}$ and $\displaystyle \frac{1}{11}$.

Insert two arith­metic means between 2 and 11: $\displaystyle d = \frac{11 - 2}{3} = 3$, giv­ing 5 and 8. So the har­monic means are $\displaystyle \frac{1}{5}$ and $\displaystyle \frac{1}{8}$, and $\displaystyle \frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}$ is an H.P.

A.M., G.M. and H.M. together#

For two pos­i­tive num­bers $a$ and $b$, let $\displaystyle A = \frac{a + b}{2}$, $G = \sqrt{ab}$ and $\displaystyle H = \frac{2ab}{a + b}$. Then:

1. $G^2 = AH$. Indeed $\displaystyle AH = \frac{a + b}{2} \cdot \frac{2ab}{a + b} = ab = G^2$. So $G$ is the geo­met­ric mean of $A$ and $H$, and $A$, $G$, $H$ are in G.P.

2. $A \ge G \ge H$, with equal­ity only when $a = b$. We proved $A \ge G$ in the les­son on geo­met­ric means. Since $G^2 = AH$, $\displaystyle \frac{G}{H} = \frac{A}{G} \ge 1$, so $G \ge H$ as well.

Exam­ple 5#

Find the A.M., G.M. and H.M. of 4 and 12, and check both results.

$A = 8$, $G = \sqrt{48} = 4\sqrt{3}$ (about 6.93), $H = 6$. Then $AH = 48 = G^2$, and $8 \ge 6.93 \ge 6$.

Exam­ple 6#

The A.M. of two num­bers is 25 and their H.M. is 16. Find the num­bers.

$G^2 = AH = 400$, so the prod­uct is 400, and the sum is $2A = 50$. The num­bers are the roots of $x^2 - 50x + 400 = 0$, that is $(x - 10)(x - 40) = 0$. They are 10 and 40. Check: $\displaystyle \frac{2 \times 10 \times 40}{50} = 16$.

Prac­tice#

1. Show that $\displaystyle \frac{1}{3}, \frac{1}{7}, \frac{1}{11}, \frac{1}{15}$ is an H.P., and write its next term.
2. Find the 8th term of the H.P. $\displaystyle 1, \frac{1}{4}, \frac{1}{7}, \ldots$
3. Find the 6th term of the H.P. $6, 3, 2, \ldots$
4. If $2$, $b$, $6$ are in H.P., find $b$.
5. Check Prop­erty 2 for the H.P. $2, 3, 6$.
6. Find the H.M. of 6 and 18.
7. Insert three har­monic means between $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{1}{23}$.
8. A cyclist rides to school at 12 km/h and back home at 20 km/h along the same road. Find the aver­age speed for the whole jour­ney.
9. Find the A.M., G.M. and H.M. of 9 and 16, and ver­ify that $G^2 = AH$.
10. Can two pos­i­tive num­bers have G.M. 12 and H.M. $\displaystyle \frac{72}{5}$? Give a rea­son.

Answers#