NDA Maths · Sequence & Series

Harmonic Progressions and the Three Means

A harmonic progression is just an AP turned upside down — its reciprocals are in AP — and it comes packaged with the three classical means AM, GM, HM and the inequality that orders them.

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Why this matters

A small but conceptually central subtopic — five PYQs, but the AM-GM-HM machinery underpins the harder interrelating-progressions questions and the compound-trick traps. The one rule to burn in: an HP problem is solved by flipping to its AP of reciprocals. There is no "sum of an HP" formula — that is the trap.

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Concept 1 of 3: Harmonic progression — flip to the reciprocal AP

Numbers are in harmonic progression when their reciprocals are in arithmetic progression. That one sentence is the whole method: never work with an HP directly — take reciprocals, solve the AP, then flip back. There is deliberately no closed formula for the sum of an HP.

Definition

$a_1, a_2, a_3, \ldots$ are in harmonic progression (HP) $\iff \tfrac{1}{a_1}, \tfrac{1}{a_2}, \tfrac{1}{a_3}, \ldots$ are in AP. So the nth term of an HP is $a_n = \dfrac{1}{a + (n-1)d}$ , where $a, d$ are the first term and common difference of the reciprocal AP. The three-term condition: $a, b, c$ are in HP $\iff b = \dfrac{2ac}{a+c}$ (equivalently $\tfrac{2}{b} = \tfrac1a + \tfrac1c$ ).

HP nth term and three-term condition

a_n = \frac{1}{a + (n-1)d}, \qquad b = \frac{2ac}{a+c}\ \text{(for } a,b,c \text{ in HP)}

Worked example

Find the 4th term of the HP

\tfrac12, \tfrac15, \tfrac18, \ldots

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2023 · Q10Moderate

Example 1 · Sequence & Series · Harmonic Progressions and the Three Means

(a+b),\ 2b,\ (b+c)

are in HP, then which one of the following is correct?

You cannot average HP terms directly

The middle term of three numbers in HP is NOT

\tfrac{a+c}{2}

— that is the arithmetic mean. The HP middle is the harmonic mean

b = \dfrac{2ac}{a+c}

(equivalently,

\tfrac1a, \tfrac1b, \tfrac1c

are in AP). For

a = 2, c = 6

the HP middle is

\dfrac{2\cdot 2\cdot 6}{8} = 3

, not

\tfrac{2+6}{2} = 4

. Flip to reciprocals first, always.

Drill 2 more on harmonic progression — flip to the reciprocal ap

Concept 2 of 3: AM, GM, HM and the inequality that orders them

Two positive numbers have three classical averages. The arithmetic mean is the everyday average; the geometric mean is the square root of the product; the harmonic mean is the reciprocal-average. For any two unequal positive numbers they line up in a fixed order, AM

>

>

HM, and the GM is always exactly the geometric mean of the other two.

Definition

For positive $a, b$ :

AM $= \dfrac{a+b}{2}$ , GM $= \sqrt{ab}$ , HM $= \dfrac{2ab}{a+b}$ .
Ordering: $\text{AM} \ge \text{GM} \ge \text{HM}$ , with equality only when $a = b$ .
Key identity: $\text{GM}^2 = \text{AM} \times \text{HM}$ — so the GM is the geometric mean of the AM and HM.
From a ratio of means back to the numbers: if $\text{AM} : \text{GM} = p : q$ , put $t = \sqrt{a/b}$ so that $\dfrac{a+b}{2\sqrt{ab}} = \dfrac{t^2 + 1}{2t} = \dfrac pq$ ; solve the quadratic $q\,t^2 - 2p\,t + q = 0$ for $t$ , and $a : b = t^2 : 1$ . E.g. $\text{AM}:\text{GM} = 5:4$ gives $4t^2 - 10t + 4 = 0$ , $t = 2$ (or $\tfrac12$ ), so $a : b = 4 : 1$ . The two roots are reciprocals, i.e. the same pair the other way round.

The three means and their relation

\text{AM} = \frac{a+b}{2},\quad \text{GM} = \sqrt{ab},\quad \text{HM} = \frac{2ab}{a+b}, \qquad \text{GM}^2 = \text{AM}\cdot\text{HM}

Worked example

Find the AM, GM, and HM of

4

and

16

, and verify

\text{GM}^2 = \text{AM}\cdot\text{HM}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2018 · Q9Moderate

Example 2 · Sequence & Series · Harmonic Progressions and the Three Means

If the ratio of AM to GM of two positive numbers a and b is

5:3

, then

a:b

is equal to

AM ≥ GM ≥ HM only for positives

The ordering and the equality-when-equal rule need

a, b > 0

. A common NDA setup gives AM and GM and asks for HM — reach straight for

\text{HM} = \tfrac{\text{GM}^2}{\text{AM}}

rather than solving for

a, b

first.

Don't swap the three mean formulas

Keep them straight:

\text{AM} = \dfrac{a+b}{2}

(sum over 2),

\text{GM} = \sqrt{ab}

(root of the product), and

\text{HM} = \dfrac{2ab}{a+b}

(twice the product over the sum). The classic slip is using

\dfrac{a+b}{2ab}

— that is the reciprocal of the HM, not the HM. For

3

and

6

\text{HM} = \dfrac{2\cdot 18}{9} = 4

, not

\dfrac{9}{36} = \tfrac14

The identity is $\text{GM}^2 = \text{AM}\cdot\text{HM}$

The geometric mean is the geometric mean of the OTHER two:

\text{GM}^2 = \text{AM}\cdot\text{HM}

, so

\text{GM} = \sqrt{\text{AM}\cdot\text{HM}}

. It is NOT

\text{AM}^2 = \text{GM}\cdot\text{HM}

and the three means are in GP (not AP). Given AM

= 9

, GM

= 6

\text{HM} = \dfrac{\text{GM}^2}{\text{AM}} = \dfrac{36}{9} = 4

, not

\dfrac{\text{AM}^2}{\text{GM}} = \dfrac{81}{6}

Concept 3 of 3: Harmonic mean of several numbers

The harmonic mean of a set is the count divided by the sum of reciprocals — it is the right average when the quantities are rates (think average speed over equal distances). For two numbers it reduces to the familiar

\tfrac{2ab}{a+b}

Definition

The harmonic mean of $n$ positive numbers $x_1, \ldots, x_n$ is

\text{HM} = \frac{n}{\dfrac{1}{x_1} + \dfrac{1}{x_2} + \cdots + \dfrac{1}{x_n}}.

Equivalently,

\tfrac{1}{\text{HM}}

is the arithmetic mean of the reciprocals.

Harmonic mean of n numbers

\text{HM} = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}

Worked example

Find the harmonic mean of

2, 3,

and

6

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2023 · Q113Hard

Example 3 · Sequence & Series · Harmonic Progressions and the Three Means

H

is the Harmonic Mean of

\binom{10}{4}

\binom{10}{5}

, and

\binom{10}{6}

, then what is the value of

\frac{270}{H}

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Harmonic progression — flip to the reciprocal AP
HP nth term and three-term condition
$a_n = \frac{1}{a + (n-1)d}, \qquad b = \frac{2ac}{a+c}\ \text{(for } a,b,c \text{ in HP)}$
AM, GM, HM and the inequality that orders them
The three means and their relation
$\text{AM} = \frac{a+b}{2},\quad \text{GM} = \sqrt{ab},\quad \text{HM} = \frac{2ab}{a+b}, \qquad \text{GM}^2 = \text{AM}\cdot\text{HM}$
Harmonic mean of several numbers
Harmonic mean of n numbers
$\text{HM} = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}}$

Watch out for (4)

You cannot average HP terms directly
→ Harmonic progression — flip to the reciprocal AP
AM ≥ GM ≥ HM only for positives
→ AM, GM, HM and the inequality that orders them
Don't swap the three mean formulas
→ AM, GM, HM and the inequality that orders them
The identity is $\text{GM}^2 = \text{AM}\cdot\text{HM}$
→ AM, GM, HM and the inequality that orders them

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