Sequences & Progressions

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Harmonic Progressions & AGP

⏱ 4 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

A harmonic progression (HP) is a list whose reciprocals form an AP, so every HP question is solved on the reciprocals. Two related tools sit beside it: the AM-GM-HM trio for two positive numbers, and the arithmetico-geometric series, where an AP and a GP are multiplied term by term.

What a harmonic progression is

A harmonic progression (HP) is a list whose reciprocals form an AP.

1/2, 1/5, 1/8, 1/11 is an HP. Turn it upside down: 2, 5, 8, 11 — an AP with $a = 2$ and $d = 3$ .

The HP terms themselves have no common difference. Never look for one.

Rule: Every HP question is solved on the reciprocals. Flip first, work in the AP, flip back at the end.

Working with an HP

The $n$ -th term of the HP is the reciprocal of the AP's $n$ -th term:

a_n = \frac{1}{a + (n-1)d}

Here $a$ and $d$ belong to the AP of reciprocals.

Which term of 1/2, 1/5, 1/8, ... is 1/32? Solve $2 + (n-1)3 = 32$ , so $n = 11$ : the 11th term.

Which term of 1/5, 1/8, 1/11, ... is 1/29? Solve $5 + (n-1)3 = 29$ , so $n = 9$ .

Given two HP terms, write the two reciprocals as AP terms. If the 4th term is 1/10 and the 10th is 1/28, then $a_4 = 10$ and $a_{10} = 28$ , so $d = 3$ and $a = 1$ . The 15th HP term is $\dfrac{1}{1 + 14 \times 3} = \dfrac{1}{43}$ .

Tip: In $\dfrac{1}{a + (n-1)d}$ the whole expression sits under the 1. Writing $\dfrac{1}{a} + (n-1)\dfrac{1}{d}$ is a different list.

The AM-GM-HM trio

For two positive numbers $x$ and $y$ : AM $= \dfrac{x+y}{2}$ , GM $= \sqrt{xy}$ , HM $= \dfrac{2xy}{x+y}$ .

They always line up in the order AM $\ge$ GM $\ge$ HM.

The three are tied by one clean relation:

AM \times HM = GM^2

The AM of two numbers is 10 and the GM is 8, so the HM is $\dfrac{64}{10} = 6.4$ .

The relation also finds the numbers. AM is 13 and GM is 12: sum 26, product 144, so the numbers are 8 and 18.

Note: For two numbers, the HM is the reciprocal of the AM of the reciprocals — the HP idea again.

What an AGP looks like

An arithmetico-geometric series (AGP) multiplies two patterns: the coefficients grow by addition, the powers grow by multiplication.

1 + 2 \cdot 2 + 3 \cdot 4 + 4 \cdot 8 + 5 \cdot 16

The numbers 1, 2, 3, 4, 5 form an AP; the powers of 2 form a GP. Neither plain sum formula works.

The AGP subtraction method

Multiply the whole series by $r$ and write it below $S$ , with equal powers lined up. Subtract.

For $S = 1 + 2 \cdot 2 + 3 \cdot 4 + 4 \cdot 8 + 5 \cdot 16 = 129$ :

$S - 2S = 1 + 2 + 4 + 8 + 16 - 160 = -129$ , so $S = 129$ .

The coefficients drop by one, leaving a small GP minus one far end. That leftover sum is easy.

Tip: Line up the powers before subtracting. A misaligned shift is the classic error here.

Infinite AGP sums

When $|r| < 1$ , the finite method settles into a formula. For $a + (a+d)r + (a+2d)r^2 + \cdots$ :

S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2}

$1 + \dfrac{2}{3} + \dfrac{3}{9} + \dfrac{4}{27} + \cdots$ : here $a = 1$ , $d = 1$ , $r = \dfrac{1}{3}$ . Sum $= \dfrac{3}{2} + \dfrac{1/3}{4/9} = \dfrac{3}{2} + \dfrac{3}{4} = \dfrac{9}{4}$ .

Watch: The formula needs $|r| < 1$ . For $r = 2$ the terms grow and no sum exists.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common4 practice Q

Which term of an HP is it?

How to spot it:

A fraction list like 1/2, 1/5, 1/8 is given and the position of one term is asked.

Method

Flip the list into its AP.
Solve $a + (n-1)d =$ target.
Report the term number $n$ .

Why it works:

The reciprocals are an AP, so position questions become ordinary AP equations.

Try this

Which term of the HP 1/2, 1/5, 1/8, ... is 1/32?

Show solution

AP of reciprocals: 2, 5, 8, ...
$2 + (n-1)3 = 32$ , so $n = 11$ .

Answer

11th term

Type 2common2 practice Q

Find an HP term from two given HP terms

How to spot it:

Two terms of an HP are given (say the 4th and the 10th) and another term is asked.

Method

Write the two reciprocals as AP terms.
Get $d$ , then $a$ , from the pair.
Compute the AP term and flip it back.

Why it works:

Two AP terms fix $a$ and $d$ , and every other term follows from them.

Try this

The 4th term of an HP is 1/10 and the 10th term is 1/28. Find the 15th term.

Show solution

$a_4 = 10$ , $a_{10} = 28$ : $d = 3$ , $a = 1$ .
15th AP term $= 1 + 14 \times 3 = 43$ .

Answer

1/43

Type 3common2 practice Q

AM, GM, HM of two numbers

How to spot it:

AM, GM or HM of two positive numbers is given (two of them), and the third — or the numbers — are asked.

AM \times HM = GM^2

Method

Use $AM \times HM = GM^2$ for the missing average.
AM gives the sum; GM gives the product.
Solve the quadratic for the two numbers.

Why it works:

AM fixes the sum, GM fixes the product, and two numbers with a given sum and product are unique up to order.

Try this

The AM of two positive numbers is 13 and their GM is 12. Find the numbers.

Show solution

Sum $= 26$ , product $= 144$ .
$t^2 - 26t + 144 = 0$ gives $t = 8, 18$ .

Answer

8 and 18

Type 4occasional2 practice Q

AGP: finite sum by subtraction

How to spot it:

Terms like 1 + 2·2 + 3·4 + 4·8 — a counting part times a powering part.

Method

Write $S$ and $rS$ with powers aligned.
Subtract: coefficients drop by 1.
Sum the leftover small GP and solve for $S$ .

Why it works:

The shift by one power turns each AGP term into a single GP term plus a constant.

Try this

Find the sum 1 + 2×2 + 3×4 + 4×8 + 5×16.

Show solution

$S - 2S = 1 + 2 + 4 + 8 + 16 - 160 = -129$ .
So $S = 129$ .

Answer

129

Type 5occasional2 practice Q

Infinite AGP sum

How to spot it:

An endless AGP like 1 + 2/3 + 3/9 + 4/27 + ... with $|r| < 1$ .

\frac{a}{1-r} + \frac{dr}{(1-r)^2}

Method

Confirm $|r| < 1$ .
Name $a$ , $d$ , $r$ .
Apply the two-part formula.

Why it works:

As the powers die out, the subtraction method leaves exactly $\dfrac{a}{1-r} + \dfrac{dr}{(1-r)^2}$ .

Try this

Find the sum 1 + 2/3 + 3/9 + 4/27 + ...

Show solution

$a = 1$ , $d = 1$ , $r = \dfrac{1}{3}$ .
$\dfrac{3}{2} + \dfrac{1/3}{4/9} = \dfrac{3}{2} + \dfrac{3}{4}$ .

Answer

9/4

Formula sheet

nth term of an HP

\frac{1}{a + (n-1)d}

$a$, $d$ come from the AP of reciprocals.

HM of two numbers

\frac{2ab}{a + b}

The reciprocal of the AM of the reciprocals.

AM x HM = GM squared

AM \times HM = GM^2

For positive numbers, AM $\ge$ GM $\ge$ HM.

Infinite AGP sum

\frac{a}{1-r} + \frac{dr}{(1-r)^2}

For $|r| < 1$, series $a + (a+d)r + (a+2d)r^2 + \cdots$

Shortcuts that save time

⚡ Flip every HP into its AP

Reciprocal the list, do all the work with AP tools, reciprocal the final answer. This solves every HP question.

Example

Which term of the HP 1/2, 1/5, 1/8, ... is 1/32?

Show solution

Reciprocals: 2, 5, 8, ... ( $d = 3$ ).
$2 + (n-1)3 = 32$ gives $n = 11$ .

Answer

11th term

⚡ AM x HM = GM squared

One relation replaces a page of algebra. Any two of AM, GM, HM give the third.

Example

The AM of two numbers is 10 and their GM is 8. Find the HM.

Show solution

$AM \times HM = GM^2$ .
$HM = \dfrac{8^2}{10} = 6.4$ .

Answer

6.4

⚡ AGP: multiply by r and subtract

Write $rS$ below $S$ with the powers aligned. Subtracting kills the middle and leaves a small GP minus one end.

Example

Find the sum 1 + 2×2 + 3×4 + 4×8 + 5×16.

Show solution

Line up $S$ and $2S$ with equal powers of 2.
$S - 2S = 1 + 2 + 4 + 8 + 16 - 160 = -129$ .
So $S = 129$ .

Answer

129

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Looking for a common difference in an HP — the HP terms have none; their reciprocals do.

Mistake 02

Writing the HP nth term as $\dfrac{1}{a} + (n-1)\dfrac{1}{d}$ — the whole denominator changes: $\dfrac{1}{a + (n-1)d}$ .

Mistake 03

Using the infinite AGP sum without checking $|r| < 1$ — the formula fails for $r = 2$ or larger.

Mistake 04

Writing AM $\le$ GM $\le$ HM — for positive numbers the order is AM $\ge$ GM $\ge$ HM.

Mistake 05

Subtracting the shifted AGP series with powers out of line — align matching powers before subtracting.

Mistake 06

Reporting an HM larger than the AM — that is impossible for positive numbers, so recheck the working.

Quick revision

Read this the night before the exam.

HP: reciprocals form an AP — always flip first.
HP nth term: $\dfrac{1}{a + (n-1)d}$ .
AM $\times$ HM $=$ GM $^2$ ; AM $\ge$ GM $\ge$ HM.
HM of $a, b$ : $\dfrac{2ab}{a+b}$ .
AGP: multiply $S$ by $r$ , align, subtract.
Infinite AGP: $\dfrac{a}{1-r} + \dfrac{dr}{(1-r)^2}$ for $|r| < 1$ .

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 10 min · wrong answers go to your mistake notebook automatically.

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