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.

Concept 1 of 3

Harmonic progression — flip to the reciprocal AP

Intuition

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

Take reciprocals: $2, 5, 8, \ldots$ — an AP with $a = 2,\ d = 3$ .
4th term of the AP: $2 + 3(3) = 11$ .
Flip back: the 4th term of the HP is $\tfrac{1}{11}$ .

Answer:

\tfrac{1}{11}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Sequence & SeriesMODERATE

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

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

[Q10 · Sep · 2023]

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

Intuition

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.

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}

$\text{AM} = \dfrac{4 + 16}{2} = 10$ .
$\text{GM} = \sqrt{4 \times 16} = \sqrt{64} = 8$ .
$\text{HM} = \dfrac{2 \times 4 \times 16}{4 + 16} = \dfrac{128}{20} = 6.4$ .
Check: $\text{AM}\cdot\text{HM} = 10 \times 6.4 = 64 = 8^2 = \text{GM}^2$ . ✓ And $10 > 8 > 6.4$ .

Answer:AM

= 10

, GM

= 8

, HM

= 6.4

; the identity holds.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Sequence & SeriesMODERATE

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

5:3

, then

a:b

is equal to

[Q9 · Apr · 2018]

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

Intuition

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

Sum of reciprocals: $\tfrac12 + \tfrac13 + \tfrac16 = \tfrac{3 + 2 + 1}{6} = 1$ .
$\text{HM} = \dfrac{n}{\sum 1/x_i} = \dfrac{3}{1}$ .

Answer:

\text{HM} = 3

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Sequence & SeriesHARD

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}

[Q113 · Sep · 2023]

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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