Standard progressions

This chapter rigorously examines the foundational concepts of standard progressions, including arithmetic, geometric, and harmonic sequences, alongside their associated means. Mastery of these topics is crucial for the CMI examination, as they form the bedrock for advanced algebraic problem-solving and quantitative analysis.

---

Chapter Contents

| Topic |

|---|-------| | 1 | Arithmetic progression | | 2 | Geometric progression | | 3 | Harmonic progression | | 4 | Means and their relations |

---

We begin with Arithmetic progression.

Part 1: Arithmetic progression

Arithmetic Progression

Overview

An arithmetic progression (AP) is a sequence in which consecutive terms differ by a fixed constant. In CMI-style questions, AP is tested through term formulas, sums, insertion of means, parameter conditions, and algebraic manipulation of indexed terms. The main strength is to move smoothly between term form, difference form, and sum form. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

identify and construct an arithmetic progression

find the $n$ th term and the common difference

compute sums of the first $n$ terms efficiently

solve condition-based problems involving AP terms

use arithmetic means and symmetry of APs correctly

---

Core Definition

📖 Arithmetic Progression

A sequence
$\qquad a_1, a_2, a_3, \dots$
is called an arithmetic progression if the difference between consecutive terms is constant.

If the first term is $a$ and the common difference is $d$ , then the AP is

$\qquad a,\ a+d,\ a+2d,\ a+3d,\dots$

📐 General Term

If the first term is $a$ and the common difference is $d$ , then the $n$ th term is

$\qquad a_n = a + (n-1)d$

---

Standard Forms

📐 Useful Representations of an AP

First-term form:

\qquad a,\ a+d,\ a+2d,\dots

Centered form with three consecutive terms:

\qquad a-d,\ a,\ a+d

Four consecutive terms:

\qquad a-3d,\ a-d,\ a+d,\ a+3d

These centered forms are very useful in symmetric condition problems.

---

Common Difference and Reconstruction

📐 Recovering the AP

If two terms are known:

from $a_m$ and $a_n$ ,

\qquad a_n-a_m=(n-m)d

So,

\qquad d=\dfrac{a_n-a_m}{n-m}

Then,

\qquad a=a_m-(m-1)d

---

Sum of Terms

📐 Sum of First

n

Terms

If an AP has first term $a$ and common difference $d$ , then

$\qquad S_n=\dfrac{n}{2}\big[2a+(n-1)d\big]$

Equivalently, using first and last terms:

$\qquad S_n=\dfrac{n}{2}(a_1+a_n)$

💡 Pairing Trick

In an AP, terms from the beginning and end pair to the same sum:

$\qquad a_1+a_n = a_2+a_{n-1} = a_3+a_{n-2}$

This is the fastest way to derive and remember the sum formula.

---

Arithmetic Mean

📐 Arithmetic Mean of Two Numbers

The arithmetic mean of two numbers $x$ and $y$ is

$\qquad \dfrac{x+y}{2}$

If $x,\ A,\ y$ are in AP, then

$\qquad 2A=x+y$

📐 Inserting Arithmetic Means

If $m$ arithmetic means are inserted between $x$ and $y$ , then the full sequence has

$\qquad m+2$

terms and common difference

$\qquad d=\dfrac{y-x}{m+1}$

---

Important Properties

📐 Key AP Identities

Middle term relation:

a,b,c

are in AP, then

\qquad 2b=a+c

Equal distance terms:

\qquad a_{n-r}+a_{n+r}=2a_n

Difference of consecutive terms:

\qquad a_{n+1}-a_n=d

Sum of first $n$ natural numbers:

\qquad 1+2+\cdots+n=\dfrac{n(n+1)}{2}

Sum of first $n$ odd numbers:

\qquad 1+3+5+\cdots+(2n-1)=n^2

---

Sign and Growth Cases

❗ How AP Behaves

if $d>0$ , the AP is increasing

if $d<0$ , the AP is decreasing

if $d=0$ , all terms are equal

So many condition questions reduce to identifying the sign of

d

---

Minimal Worked Examples

Example 1 If the first term is

7

and common difference is

5

, then

\qquad a_n = 7+(n-1)5 = 5n+2

--- Example 2 Find the sum of the first

20

terms of

\qquad 3,7,11,\dots

Here,

\qquad a=3,\ d=4

So,

\qquad S_{20}=\dfrac{20}{2}[2\cdot 3+19\cdot 4]

\qquad =10(6+76)=820

---

CMI Strategy

💡 How to Attack AP Problems

first decide whether the question is about terms or sums

if specific indexed terms are given, use $a_n=a+(n-1)d$

if a sum is involved, use $S_n=\dfrac{n}{2}(a_1+a_n)$ whenever possible

in three-term AP problems, use the centered form $a-d,\ a,\ a+d$

watch for hidden AP conditions in linear equations and inserted means

in harder problems, convert every condition into equations in $a$ and $d$

---

Common Mistakes

⚠️ Avoid These Errors

❌ using $a_n=a+nd$ instead of $a+(n-1)d$

❌ confusing a term with a sum

❌ forgetting that inserted means create extra total terms

❌ using the sum formula without finding the correct last term

❌ not using symmetry when terms are equally spaced

---

Practice Questions

:::question type="MCQ" question="If the

5

th term of an AP is

18

and the common difference is

3

, then the first term is" options=["

3

","

6

","

9

","

12

"] answer="B" hint="Use

a_n=a+(n-1)d

." solution="We use

\qquad a_5=a+4d

Given

\qquad 18=a+4\cdot 3=a+12

\qquad a=6

Hence the correct option is

\boxed{B}

." ::: :::question type="NAT" question="Find the sum of the first

15

terms of the AP

4,7,10,\dots

." answer="375" hint="Use the formula for the sum of the first

n

terms." solution="Here

\qquad a=4,\ d=3,\ n=15

\qquad S_{15}=\dfrac{15}{2}[2\cdot 4 + 14\cdot 3]

\qquad =\dfrac{15}{2}(8+42)=\dfrac{15}{2}\cdot 50=375

Hence the answer is

\boxed{375}

." ::: :::question type="MSQ" question="Which of the following statements are true for an arithmetic progression?" options=["If

a,b,c

are consecutive terms of an AP, then

2b=a+c

.","For an AP,

a_{n-r}+a_{n+r}=2a_n

.","The sum of the first

n

terms is always

na_n

.","If the common difference is zero, then all terms are equal."] answer="A,B,D" hint="Use the general term and test each statement." solution="1. True. Consecutive terms in AP satisfy

\qquad a,b,c = x-d,\ x,\ x+d

\qquad 2b=a+c

True. Using

\qquad a_n=a+(n-1)d

, we get

\qquad a_{n-r}=a+(n-r-1)d

and

\qquad a_{n+r}=a+(n+r-1)d

Adding gives

\qquad 2a+2(n-1)d = 2a_n

False. The correct formula is

\qquad S_n=\dfrac{n}{2}(a_1+a_n)

, not always

na_n

True. If

d=0

, then every term equals the first term.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="If the

3

rd term of an AP is

11

and the

9

th term is

29

, find the first term, the common difference, and the sum of the first

20

terms." answer="

a=5,\\ d=3,\\ S_{20}=670

" hint="Form two equations using the term formula." solution="Let the first term be

a

and common difference be

d

. Given:

\qquad a_3=a+2d=11

\qquad a_9=a+8d=29

Subtracting,

\qquad 6d=18

\qquad d=3

Substitute into

\qquad a+2d=11

\qquad a+6=11

\qquad a=5

Now compute the sum of the first

20

terms:

\qquad S_{20}=\dfrac{20}{2}[2a+19d]

\qquad =10[2\cdot 5 + 19\cdot 3]

\qquad =10(10+57)=670

Hence

\qquad \boxed{a=5,\ d=3,\ S_{20}=670}

" ::: ---

Summary

❗ Key Takeaways for CMI

An AP is determined by its first term $a$ and common difference $d$ .

The $n$ th term is $\qquad a_n=a+(n-1)d$ .

The sum of the first $n$ terms is $\qquad S_n=\dfrac{n}{2}[2a+(n-1)d]$ .

If $a,b,c$ are in AP, then $\qquad 2b=a+c$ .

Symmetry of equally spaced terms is a major simplification tool.

Many AP questions reduce to solving linear equations in $a$ and $d$ .

---

💡 Next Up

Proceeding to Geometric progression.

---

Part 2: Geometric progression

Geometric Progression

Overview

A geometric progression (GP) is a sequence in which consecutive terms are obtained by multiplying by a fixed nonzero constant ratio. In CMI-style questions, GP is tested through term formulas, sums, growth and decay, inserted means, and equations involving ratios or exponents. The real strength here is being comfortable with ratio thinking. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

identify and construct a geometric progression

find the $n$ th term and the common ratio

compute finite and infinite sums where valid

use geometric means and inserted means

solve parameter-based and exponent-style GP questions

---

Core Definition

📖 Geometric Progression

A sequence
$\qquad a_1,a_2,a_3,\dots$
is called a geometric progression if the ratio of each term to the previous term is constant.

If the first term is $a$ and the common ratio is $r$ , then the GP is

$\qquad a,\ ar,\ ar^2,\ ar^3,\dots$

📐 General Term

If the first term is $a$ and the common ratio is $r$ , then the $n$ th term is

$\qquad a_n = ar^{n-1}$

---

Ratio Recovery

📐 Finding the Ratio

If two terms are known:
$\qquad a_m=ar^{m-1}, \quad a_n=ar^{n-1}$

then

$\qquad \dfrac{a_n}{a_m}=r^{n-m}$

This is often the cleanest way to recover $r$ .

---

Geometric Mean

📐 Geometric Mean of Two Positive Numbers

If $x,\ G,\ y$ are in GP and $x,y>0$ , then

$\qquad G^2=xy$

So the geometric mean is

$\qquad G=\sqrt{xy}$

❗ Contrast with AP

AP middle term: $\qquad 2A=x+y$

GP middle term: $\qquad G^2=xy$

This distinction is tested frequently.

---

Sum of a Finite GP

📐 Sum of First

n

Terms

For a GP with first term $a$ and ratio $r\ne 1$ ,

$\qquad S_n = a\dfrac{r^n-1}{r-1}$

Equivalent form:

$\qquad S_n = a\dfrac{1-r^n}{1-r}$

If $r=1$ , then all terms are equal and

$\qquad S_n = na$

💡 Which Form to Use?

use $\qquad a\dfrac{r^n-1}{r-1}$ when $r>1$

use $\qquad a\dfrac{1-r^n}{1-r}$ when $|r|<1$

Both are algebraically the same when

r\ne 1

---

Infinite GP

📐 Sum to Infinity

If $|r|<1$ , then the infinite GP

$\qquad a+ar+ar^2+\cdots$

has sum

$\qquad S_\infty=\dfrac{a}{1-r}$

If $|r|\ge 1$ , the infinite sum does not converge.

---

Inserted Geometric Means

📐 Inserting Geometric Means

If $m$ geometric means are inserted between two nonzero numbers $x$ and $y$ , then the full GP has

$\qquad m+2$

terms.

If the common ratio is $r$ , then

$\qquad y = x r^{m+1}$

So

$\qquad r=\left(\dfrac{y}{x}\right)^{1/(m+1)}$

when the expression is meaningful in the required number system.

---

Important Properties

📐 Key GP Identities

Consecutive terms:

a,b,c

are in GP, then

\qquad b^2=ac

Equal distance terms:

\qquad a_{n-r}a_{n+r}=a_n^2

Ratio form:

\qquad \dfrac{a_{n+1}}{a_n}=r

Power sum:

\qquad 1+r+r^2+\cdots+r^{n-1}=\dfrac{r^n-1}{r-1}

for

r\ne 1

---

Behaviour by Ratio

❗ How GP Behaves

if $r>1$ and $a>0$ , terms grow

if $0<r<1$ , terms decay

if $r<0$ , signs alternate

if $r=1$ , the sequence is constant

if $r=0$ , from the second term onward everything becomes zero

Many exam questions depend on the sign or magnitude of

r

---

Minimal Worked Examples

Example 1 If first term is

3

and ratio is

2

, then

\qquad a_n=3\cdot 2^{n-1}

--- Example 2 Find the sum of the first

6

terms of

\qquad 5,10,20,\dots

Here

\qquad a=5,\ r=2

\qquad S_6=5\cdot \dfrac{2^6-1}{2-1}=5(64-1)=315

---

CMI Strategy

💡 How to Attack GP Problems

first decide whether the problem is about terms, ratios, or sums

write everything using $a_n=ar^{n-1}$

divide terms to eliminate $a$

for sum questions, first check whether $r=1$

for infinite sums, always check $|r|<1$

in three-term GP problems, use the form $\dfrac{a}{r},\ a,\ ar$ or directly use $b^2=ac$

---

Common Mistakes

⚠️ Avoid These Errors

❌ confusing common difference with common ratio

❌ forgetting the special case $r=1$

❌ using sum to infinity when $|r|\ge 1$

❌ using $b=\dfrac{x+y}{2}$ instead of $b^2=xy$ for GP

❌ sign mistakes when $r<0$

---

Practice Questions

:::question type="MCQ" question="If the first term of a GP is

2

and the common ratio is

3

, then the

5

th term is" options=["

54

","

81

","

162

","

243

"] answer="C" hint="Use

a_n=ar^{n-1}

." solution="We use

\qquad a_n=ar^{n-1}

\qquad a_5 = 2\cdot 3^{4}=2\cdot 81=162

Hence the correct option is

\boxed{C}

." ::: :::question type="NAT" question="Find the sum of the first

7

terms of the GP

3,6,12,\dots

." answer="381" hint="Use the finite GP sum formula." solution="Here

\qquad a=3,\ r=2,\ n=7

\qquad S_7 = 3\cdot \dfrac{2^7-1}{2-1}

\qquad =3(128-1)=3\cdot 127=381

Hence the answer is

\boxed{381}

." ::: :::question type="MSQ" question="Which of the following statements are true for a geometric progression?" options=["If

a,b,c

are consecutive terms of a GP, then

b^2=ac

.","For a GP,

a_{n-r}a_{n+r}=a_n^2

.","The sum to infinity exists for every GP with nonzero first term.","If

r=1

, then all terms of the GP are equal."] answer="A,B,D" hint="Use the term formula and the convergence condition for infinite sums." solution="1. True. If

a,b,c

are consecutive terms of a GP, then they are of the form

\qquad \dfrac{x}{r},\ x,\ xr

\qquad x^2=\dfrac{x}{r}\cdot xr

Hence

\qquad b^2=ac

True. Using

\qquad a_n=ar^{n-1}

, we get

\qquad a_{n-r}=ar^{n-r-1}

and

\qquad a_{n+r}=ar^{n+r-1}

Their product is

\qquad a^2r^{2n-2}=a_n^2

False. Sum to infinity exists only when

\qquad |r|<1

True. If

r=1

, every term equals the first term.

Hence the correct answer is

\boxed{A,B,D}

." ::: :::question type="SUB" question="The

3

rd term of a GP is

12

and the

6

th term is

96

. Find the first term, the common ratio, and the sum of the first

6

terms." answer="

a=3,\\ r=2,\\ S_6=189

" hint="Use ratios to find

r

, then recover

a

." solution="Let the first term be

a

and common ratio be

r

. Given:

\qquad a_3=ar^2=12

\qquad a_6=ar^5=96

Divide the second by the first:

\qquad \dfrac{ar^5}{ar^2}=r^3=\dfrac{96}{12}=8

\qquad r=2

Now use

\qquad ar^2=12

\qquad 4a=12

\qquad a=3

Now compute the sum of the first

6

terms:

\qquad S_6 = a\dfrac{r^6-1}{r-1}

\qquad =3\cdot \dfrac{2^6-1}{2-1}

\qquad =3(64-1)=189

Hence

\qquad \boxed{a=3,\ r=2,\ S_6=189}

" ::: ---

Summary

❗ Key Takeaways for CMI

A GP is determined by its first term $a$ and common ratio $r$ .

The $n$ th term is $\qquad a_n=ar^{n-1}$ .

The sum of the first $n$ terms is $\qquad S_n=a\dfrac{r^n-1}{r-1}$ for $r\ne 1$ .

If $a,b,c$ are in GP, then $\qquad b^2=ac$ .

Infinite GP sum exists only when $\qquad |r|<1$ .

Ratio thinking is the fastest route in most GP problems.

---

💡 Next Up

Proceeding to Harmonic progression.

---

Part 3: Harmonic progression

Harmonic Progression

Overview

A harmonic progression is less direct than an arithmetic progression or a geometric progression. Instead of looking at the terms themselves, we look at their reciprocals. This makes HP a very important bridge topic: many questions become easy only after converting the problem into an AP of reciprocals. In CMI-style algebra, HP is usually tested through definition, insertion of harmonic means, relations with AP, algebraic manipulation, and conversion tricks rather than by memorising isolated formulas. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

define a harmonic progression correctly

convert HP questions into AP questions using reciprocals

find terms and means in a harmonic progression

derive and use the harmonic mean formula

handle algebraic manipulations involving three or more terms in HP

---

Definition

📖 Harmonic Progression

A sequence

$\qquad a_1,\ a_2,\ a_3,\dots$

of nonzero numbers is said to be in harmonic progression if the sequence of reciprocals

$\qquad \dfrac1{a_1},\ \dfrac1{a_2},\ \dfrac1{a_3},\dots$

is in arithmetic progression.

❗ Immediate Consequence

If $a,b,c$ are in HP, then

$\qquad \dfrac1a,\ \dfrac1b,\ \dfrac1c$

are in AP.

---

Core Criterion

📐 Three Numbers in HP

If $a,b,c$ are in HP, then

$\qquad \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$

This is just the AP condition on reciprocals:

$\qquad 2\left(\dfrac1b\right)=\dfrac1a+\dfrac1c$

📐 Equivalent Form

From

$\qquad \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$

we get

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

so

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

Thus, if $a$ and $c$ are the first and third terms of an HP, then the middle term is

$\qquad \boxed{\dfrac{2ac}{a+c}}$

---

Harmonic Mean

📖 Harmonic Mean of Two Numbers

The harmonic mean of two nonzero numbers $a$ and $b$ is the number $H$ such that

$\qquad a,\ H,\ b$

are in harmonic progression.

📐 Formula for Harmonic Mean

If $H$ is the harmonic mean of $a$ and $b$ , then

$\qquad \dfrac{2}{H}=\dfrac{1}{a}+\dfrac{1}{b}$

Hence

$\qquad H=\dfrac{2ab}{a+b}$

⚠️ Restriction

The harmonic mean formula uses reciprocals, so $a$ and $b$ must be nonzero.

---

General Form of an HP

📐 HP from an AP of Reciprocals

If

$\qquad \dfrac1{a_n}=A+(n-1)d$

for some constants $A,d$ , then

$\qquad a_n=\dfrac{1}{A+(n-1)d}$

So a general harmonic progression is often written as

$\qquad \dfrac1a,\ \dfrac1{a+d},\ \dfrac1{a+2d},\dots$

when the reciprocals form an AP.

---

Useful Relations

📐 Key Identities

If $a,b,c$ are in HP, then:

$\qquad \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$

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

$\qquad \dfrac{a+c}{2} \ge b$ for positive $a,c$

because harmonic mean is at most arithmetic mean

if $a,b,c$ are positive and in HP, then $b^2 \le ac$ need not hold in HP itself, but

\dfrac1a,\dfrac1b,\dfrac1c

are in AP, so many AP identities can be used after reciprocation

---

HP and AP Connection

💡 Main Solving Principle

Almost every HP problem should be converted into an AP problem.

If
$\qquad a_1,a_2,a_3,\dots$
are in HP, define
$\qquad u_n=\dfrac1{a_n}$

Then
$\qquad u_1,u_2,u_3,\dots$
are in AP.

Solve the AP problem first, then convert back.

---

Minimal Worked Examples

Example 1 Find the harmonic mean between

4

and

12

. Using the formula,

\qquad H=\dfrac{2\cdot 4 \cdot 12}{4+12}

\qquad =\dfrac{96}{16}=6

So the harmonic mean is

\boxed{6}

. --- Example 2 If

a,6,12

are in HP, find

a

. Since the reciprocals are in AP,

\qquad \dfrac{2}{6}=\dfrac{1}{a}+\dfrac{1}{12}

\qquad \dfrac13=\dfrac1a+\dfrac1{12}

\qquad \dfrac1a=\dfrac13-\dfrac1{12}=\dfrac4{12}-\dfrac1{12}=\dfrac3{12}=\dfrac14

\qquad a=4

Hence the answer is

\boxed{4}

. ---

Insertion of Harmonic Means

📐 One Harmonic Mean

If one harmonic mean is inserted between $a$ and $b$ , it is

$\qquad \dfrac{2ab}{a+b}$

📐 Several Harmonic Means

If $n$ harmonic means are inserted between $a$ and $b$ , then the reciprocals form an AP with first term $\dfrac1a$ and last term $\dfrac1b$ .

So the common difference is

$\qquad d=\dfrac{\frac1b-\frac1a}{n+1}$

and the inserted means are obtained by taking reciprocals of the intermediate AP terms.

---

Sign and Domain Observations

❗ Be Careful with Signs

HP terms must be nonzero

if all terms are positive, then standard mean inequalities apply nicely

if negative terms occur, reciprocal conversion still works, but inequality-based shortcuts must be used carefully

---

Common Errors

⚠️ Avoid These Errors

❌ treating HP like AP directly on the original terms

❌ forgetting that HP is defined through reciprocals

❌ using the harmonic mean formula when one number is zero

❌ cancelling wrongly after cross multiplication

❌ confusing the harmonic mean with the arithmetic mean

---

Strategy for CMI-Type Questions

💡 CMI Strategy

convert the HP into an AP of reciprocals immediately

use AP structure to derive the required relation

only at the end return to the original terms

for three terms, remember the shortcut

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

in insertion problems, work entirely with reciprocals first

---

Practice Questions

:::question type="MCQ" question="If

a,6,12

are in harmonic progression, then

a

is equal to" options=["

3

","

4

","

8

","

9

"] answer="B" hint="Use the reciprocals in AP condition." solution="Since

a,6,12

are in HP, we have

\qquad \dfrac{2}{6}=\dfrac{1}{a}+\dfrac{1}{12}

\qquad \dfrac13=\dfrac1a+\dfrac1{12}

Hence

\qquad \dfrac1a=\dfrac13-\dfrac1{12}=\dfrac14

Thus

\qquad a=4

Therefore the correct option is

\boxed{B}

." ::: :::question type="NAT" question="Find the harmonic mean of

3

and

6

." answer="4" hint="Use

H=\dfrac{2ab}{a+b}

." solution="The harmonic mean is

\qquad H=\dfrac{2ab}{a+b}

\qquad H=\dfrac{2\cdot 3\cdot 6}{3+6}=\dfrac{36}{9}=4

Hence the answer is

\boxed{4}

." ::: :::question type="MSQ" question="Which of the following statements are true for nonzero numbers

a,b,c

in harmonic progression?" options=["

\dfrac1a,\dfrac1b,\dfrac1c

are in arithmetic progression","

\dfrac{2}{b}=\dfrac1a+\dfrac1c

","

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

","

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

"] answer="A,B,C" hint="Translate HP into AP of reciprocals." solution="1. True. This is the definition of HP.

True. This is the AP middle-term condition on reciprocals.

True. It follows by simplifying statement 2.

False.

\dfrac{a+c}{2}

is the arithmetic mean, not the harmonic mean.

Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="If

a,b,c

are in harmonic progression, prove that

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

." answer="

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

" hint="Use the fact that the reciprocals are in AP." solution="Since

a,b,c

are in harmonic progression, their reciprocals are in arithmetic progression. Therefore,

\qquad \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}

Now combine the terms on the right:

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

Cross-multiplying gives

\qquad 2ac=b(a+c)

Hence,

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

Therefore the required result is

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

." ::: ---

Summary

❗ Key Takeaways for CMI

A sequence is in HP if and only if its reciprocals are in AP.

For three terms in HP, the middle term is

\qquad \dfrac{2ac}{a+c}

The harmonic mean of $a$ and $b$ is

\qquad \dfrac{2ab}{a+b}

Most HP problems should be solved by converting them into AP problems first.

Reciprocal thinking is the core idea of the topic.

---

💡 Next Up

Proceeding to Means and their relations.

---

Part 4: Means and their relations

Means and Their Relations

Overview

The classical means of positive numbers are among the most useful comparison tools in algebra. The arithmetic mean, geometric mean, and harmonic mean appear naturally in progressions, inequalities, optimization, and symmetric expressions. In CMI-style problems, the emphasis is not only on definitions, but on the relationships among these means and on when equality occurs. This topic is especially important because it links sequences, factorisation, inequalities, and standard progression language. ---

Learning Objectives

❗ By the End of This Topic

After studying this topic, you will be able to:

define the arithmetic, geometric, and harmonic means correctly

use the relation between means of two positive numbers

apply the inequality

\qquad \text{AM} \ge \text{GM} \ge \text{HM}

identify equality cases accurately

use means to simplify algebraic and progression-based problems

---

Definitions

📖 Arithmetic Mean

The arithmetic mean of two numbers $a$ and $b$ is

$\qquad A=\dfrac{a+b}{2}$

📖 Geometric Mean

For positive numbers $a$ and $b$ , the geometric mean is

$\qquad G=\sqrt{ab}$

📖 Harmonic Mean

For nonzero numbers $a$ and $b$ , the harmonic mean is

$\qquad H=\dfrac{2ab}{a+b}$

❗ Positivity Condition

When working with the classical relation among means, we usually assume

$\qquad a>0,\ b>0$

so that all three means are real and positive.

---

Main Relations

📐 AM, GM, HM for Two Positive Numbers

For positive numbers $a$ and $b$ :

$\qquad A=\dfrac{a+b}{2}$

$\qquad G=\sqrt{ab}$
and they satisfy

$\qquad A \ge G \ge H$

📐 Product Relation

For two positive numbers,

$\qquad AH = G^2$

Indeed,

$\qquad \dfrac{a+b}{2}\cdot \dfrac{2ab}{a+b}=ab=(\sqrt{ab})^2$

📐 Equivalent Form of HM

Since

$\qquad H=\dfrac{2ab}{a+b}$

we also have

$\qquad \dfrac{1}{H}=\dfrac12\left(\dfrac1a+\dfrac1b\right)$

So the reciprocal of the harmonic mean is the arithmetic mean of the reciprocals.

---

AM-GM Inequality

📐 AM-GM for Two Positive Numbers

For $a,b>0$ ,

$\qquad \dfrac{a+b}{2}\ge \sqrt{ab}$

Equality holds if and only if

$\qquad a=b$

💡 One Standard Proof

Since $(\sqrt{a}-\sqrt{b})^2 \ge 0$ ,

$\qquad a+b-2\sqrt{ab}\ge 0$

So

$\qquad a+b \ge 2\sqrt{ab}$

and hence

$\qquad \dfrac{a+b}{2}\ge \sqrt{ab}$

---

GM-HM Inequality

📐 GM-HM for Two Positive Numbers

For $a,b>0$ ,

$\qquad \sqrt{ab}\ge \dfrac{2ab}{a+b}$

Equality holds if and only if

$\qquad a=b$

💡 Fast Derivation

Since

$\qquad A \ge G$

and

$\qquad AH=G^2$

dividing by the positive quantity $A$ gives

$\qquad H=\dfrac{G^2}{A}\le G$

---

Combined Inequality

❗ Standard Order of Means

For positive numbers $a,b$ ,

$\qquad \boxed{\text{AM} \ge \text{GM} \ge \text{HM}}$

Equality throughout holds if and only if

$\qquad a=b$

---

Means in Progression Language

📐 Insertion of Means

Between two positive numbers $a$ and $b$ :

arithmetic mean inserted:

\qquad \dfrac{a+b}{2}

geometric mean inserted:

\qquad \sqrt{ab}

harmonic mean inserted:

\qquad \dfrac{2ab}{a+b}

📐 Three-Term Forms

If $a,b,c$ are in:

AP, then

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

GP, then

\qquad b=\sqrt{ac}

for positive

a,c

HP, then

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

---

Comparing the Means

💡 Useful Ordering Insight

For two distinct positive numbers $a$ and $b$ ,

$\qquad \text{AM} > \text{GM} > \text{HM}$

The arithmetic mean is the largest because it responds linearly to the values.
The harmonic mean is the smallest because it is controlled by reciprocals and is pulled more strongly toward the smaller number.

---

Minimal Worked Examples

Example 1 Find the AM, GM, and HM of

4

and

9

. Arithmetic mean:

\qquad A=\dfrac{4+9}{2}=\dfrac{13}{2}

Geometric mean:

\qquad G=\sqrt{4\cdot 9}=6

Harmonic mean:

\qquad H=\dfrac{2\cdot 4\cdot 9}{4+9}=\dfrac{72}{13}

\qquad \dfrac{13}{2} > 6 > \dfrac{72}{13}

--- Example 2 If the arithmetic mean of two positive numbers is

10

and their geometric mean is

8

, find the harmonic mean. Using

\qquad AH=G^2

we get

\qquad 10\cdot H = 64

\qquad H=\dfrac{64}{10}=\dfrac{32}{5}

Hence the harmonic mean is

\boxed{\dfrac{32}{5}}

. ---

Equality Condition

❗ When Are All Three Means Equal?

For positive numbers $a$ and $b$ ,

$\qquad \text{AM}=\text{GM}=\text{HM}$

if and only if

$\qquad a=b$

⚠️ Do Not Miss the Equality Case

In exam problems, many answers become incomplete if the equality condition is not stated.

---

Common Errors

⚠️ Avoid These Errors

❌ using geometric mean when numbers are not positive

❌ confusing AM with HM

❌ forgetting that $AH=G^2$ holds for two numbers

❌ writing equality in AM-GM without checking the equality condition

❌ assuming the same formulas extend unchanged to every multi-variable setting

---

Strategy for CMI-Type Questions

💡 CMI Strategy

write the three means explicitly before manipulating

for two positive numbers, remember the identity

\qquad AH=G^2

when given two means, find the third using algebra

in comparison questions, use

\qquad \text{AM} \ge \text{GM} \ge \text{HM}

always state the equality case clearly

---

Practice Questions

:::question type="MCQ" question="The harmonic mean of

4

and

12

is" options=["

6

","

8

","

9

","

10

"] answer="A" hint="Use

H=\dfrac{2ab}{a+b}

." solution="The harmonic mean is

\qquad H=\dfrac{2ab}{a+b}

\qquad H=\dfrac{2\cdot 4\cdot 12}{4+12}=\dfrac{96}{16}=6

Therefore the correct option is

\boxed{A}

." ::: :::question type="NAT" question="If the arithmetic mean and geometric mean of two positive numbers are

5

and

4

respectively, find the harmonic mean." answer="16/5" hint="Use the relation

AH=G^2

." solution="For two positive numbers,

\qquad AH=G^2

Given

\qquad A=5,\quad G=4

\qquad 5H=16

Hence

\qquad H=\dfrac{16}{5}

Therefore the answer is

\boxed{\dfrac{16}{5}}

." ::: :::question type="MSQ" question="Which of the following statements are true for two positive numbers?" options=["

\text{AM} \ge \text{GM}

","

\text{GM} \ge \text{HM}

","

AH=G^2

","

\text{HM} \ge \text{AM}

"] answer="A,B,C" hint="Recall the standard order of means." solution="For two positive numbers we know

\qquad \text{AM} \ge \text{GM} \ge \text{HM}

Also

\qquad AH=G^2

So statements 1, 2, and 3 are true, while statement 4 is false. Hence the correct answer is

\boxed{A,B,C}

." ::: :::question type="SUB" question="Prove that for two positive numbers

a

and

b

, the arithmetic mean is greater than or equal to the geometric mean." answer="

\dfrac{a+b}{2} \ge \sqrt{ab}

" hint="Start with a square that is always non-negative." solution="Since

a,b>0

, we have

\qquad (\sqrt{a}-\sqrt{b})^2 \ge 0

Expanding gives

\qquad a+b-2\sqrt{ab}\ge 0

\qquad a+b \ge 2\sqrt{ab}

Dividing both sides by

2

, we obtain

\qquad \dfrac{a+b}{2}\ge \sqrt{ab}

Equality holds if and only if

\qquad \sqrt{a}=\sqrt{b}

that is,

\qquad a=b

Therefore the result is proved:

\boxed{\dfrac{a+b}{2}\ge \sqrt{ab}}

." ::: ---

Summary

❗ Key Takeaways for CMI

For two positive numbers,

\qquad A=\dfrac{a+b}{2},\ G=\sqrt{ab},\ H=\dfrac{2ab}{a+b}

These satisfy

\qquad \text{AM} \ge \text{GM} \ge \text{HM}

Also,

\qquad AH=G^2

Equality holds if and only if the two numbers are equal.

The three means also correspond to middle terms in AP, GP, and HP respectively.

Chapter Summary

❗ Standard progressions — Key Points

Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. The $n$ -th term is $a_n = a + (n-1)d$ . The sum of $n$ terms is $S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(a + a_n)$ .

Geometric Progression (GP): A sequence where the ratio of consecutive terms is constant. The $n$ -th term is $a_n = ar^{n-1}$ . The sum of $n$ terms is $S_n = \frac{a(r^n - 1)}{r-1}$ (for $r \ne 1$ ). The sum of infinite terms is $S_\infty = \frac{a}{1-r}$ (for $|r|<1$ ).

Harmonic Progression (HP): A sequence such that the reciprocals of its terms form an Arithmetic Progression. There is no general formula for the sum of an HP.

Arithmetic Mean (AM), Geometric Mean (GM), Harmonic Mean (HM): For $n$ positive numbers $x_1, \dots, x_n$ :

$\operatorname{AM} = \frac{x_1 + \dots + x_n}{n}$

\operatorname{GM} = (x_1 \dots x_n)^{1/n}

\operatorname{HM} = \frac{n}{1/x_1 + \dots + 1/x_n}

Mean Inequalities: For any set of positive numbers, $\operatorname{AM} \ge \operatorname{GM} \ge \operatorname{HM}$ . Equality holds if and only if all numbers are equal.

Relation between AM, GM, HM for two numbers: For two positive numbers $a, b$ , the relationship $\operatorname{GM}^2 = \operatorname{AM} \times \operatorname{HM}$ holds, i.e., $(\sqrt{ab})^2 = \left(\frac{a+b}{2}\right) \left(\frac{2ab}{a+b}\right)$ .

Chapter Review Questions

:::question type="MCQ" question="If the 3rd, 6th, and 12th terms of an Arithmetic Progression are $a, b, c$ respectively, and $a, b, c$ are in Geometric Progression, then the common ratio of the GP is:" options=["1/2","1","2","4"] answer="2" hint="Let the AP have first term $A$ and common difference $D$ . Express $a, b, c$ in terms of $A$ and $D$ . Use the GP condition $b^2 = ac$ to find a relation between $A$ and $D$ , then determine the common ratio $b/a$ ." solution="Let the AP have first term $A$ and common difference $D$ .
The 3rd term is $a = A+2D$ .
The 6th term is $b = A+5D$ .
The 12th term is $c = A+11D$ .
Since $a, b, c$ are in GP, $b^2 = ac$ .
$(A+5D)^2 = (A+2D)(A+11D)$
$A^2 + 10AD + 25D^2 = A^2 + 13AD + 22D^2$
$10AD + 25D^2 = 13AD + 22D^2$
$3D^2 = 3AD \implies D^2 = AD$ .
If $D=0$ , then $A=0$ (trivial case where all terms are 0) or $A \ne 0$ (all terms $a,b,c$ are equal to $A$ , so common ratio is 1).
If $D \ne 0$ , we can divide by $D$ : $D=A$ .
Substitute $A=D$ back into the terms:
$a = A+2A = 3A$
$b = A+5A = 6A$
$c = A+11A = 12A$
The terms $a, b, c$ are $3A, 6A, 12A$ . This is a GP with common ratio $r = \frac{6A}{3A} = 2$ .

The final answer is $\boxed{\text{2}}$ "
:::

:::question type="NAT" question="The minimum value of $9\tan^2\theta + 4\cot^2\theta$ for $\theta \in (0, \pi/2)$ is:" answer="12" hint="Apply the AM-GM inequality for two positive numbers $x, y$ : $\frac{x+y}{2} \ge \sqrt{xy}$ ." solution="We want to find the minimum value of $9\tan^2\theta + 4\cot^2\theta$ .
Since $\theta \in (0, \pi/2)$ , both $\tan^2\theta$ and $\cot^2\theta$ are positive.
We can apply the AM-GM inequality: For two positive numbers $X$ and $Y$ , $\frac{X+Y}{2} \ge \sqrt{XY}$ .
Let $X = 9\tan^2\theta$ and $Y = 4\cot^2\theta$ .
$9\tan^2\theta + 4\cot^2\theta \ge 2\sqrt{(9\tan^2\theta)(4\cot^2\theta)}$
$9\tan^2\theta + 4\cot^2\theta \ge 2\sqrt{36(\tan^2\theta \cot^2\theta)}$
Since $\tan\theta \cot\theta = 1$ , we have $\tan^2\theta \cot^2\theta = 1$ .
$9\tan^2\theta + 4\cot^2\theta \ge 2\sqrt{36 \times 1}$
$9\tan^2\theta + 4\cot^2\theta \ge 2 \times 6$
$9\tan^2\theta + 4\cot^2\theta \ge 12$ .
The minimum value is 12. Equality holds when $9\tan^2\theta = 4\cot^2\theta$ , i.e., $9\tan^2\theta = \frac{4}{\tan^2\theta} \implies 9\tan^4\theta = 4 \implies \tan^2\theta = \frac{4}{9} \implies \tan\theta = \frac{2}{3}$ (since $\theta \in (0, \pi/2)$ ). This is possible.

The final answer is $\boxed{12}$ "
:::

:::question type="MCQ" question="If the 5th term of a Harmonic Progression is 1/12 and its 11th term is 1/24, what is its 16th term?" options=["1/30","1/32","1/34","1/36"] answer="1/34" hint="The reciprocals of the terms of an HP form an AP. Find the first term and common difference of this corresponding AP." solution="Let the Harmonic Progression be $h_1, h_2, h_3, \dots$ .
Its 5th term is $h_5 = 1/12$ .
Its 11th term is $h_{11} = 1/24$ .
The reciprocals of the terms of an HP form an Arithmetic Progression. Let this AP be $A_1, A_2, A_3, \dots$ with first term $A$ and common difference $D$ .
So, $A_n = 1/h_n$ .
The 5th term of the AP is $A_5 = 1/h_5 = 1/(1/12) = 12$ .
Thus, $A + (5-1)D = A + 4D = 12$ . (Equation 1)
The 11th term of the AP is $A_{11} = 1/h_{11} = 1/(1/24) = 24$ .
Thus, $A + (11-1)D = A + 10D = 24$ . (Equation 2)
Subtract Equation 1 from Equation 2:
$(A + 10D) - (A + 4D) = 24 - 12$
$6D = 12 \implies D=2$ .
Substitute $D=2$ into Equation 1:
$A + 4(2) = 12 \implies A + 8 = 12 \implies A=4$ .
We need to find the 16th term of the HP, which means finding the 16th term of the AP, $A_{16}$ .
$A_{16} = A + (16-1)D = A + 15D = 4 + 15(2) = 4 + 30 = 34$ .
The 16th term of the HP is $h_{16} = 1/A_{16} = 1/34$ .

The final answer is $\boxed{\text{1/34}}$ "
:::

:::question type="MCQ" question="If $S_n$ denotes the sum of the first $n$ terms of an arithmetic progression, and $S_{2n} = 3S_n$ , then the ratio $S_{3n} : S_n$ is:" options=["4:1","6:1","8:1","10:1"] answer="6:1" hint="Use the formula for the sum of an AP $S_k = \frac{k}{2}(2a + (k-1)d)$ . Establish a relationship between $a$ and $d$ from the given condition, then substitute to find the desired ratio." solution="Let the first term of the AP be $a$ and the common difference be $d$ .
The sum of the first $k$ terms is given by $S_k = \frac{k}{2}(2a + (k-1)d)$ .
Given $S_{2n} = 3S_n$ :
$\frac{2n}{2}(2a + (2n-1)d) = 3 \cdot \frac{n}{2}(2a + (n-1)d)$
$n(2a + (2n-1)d) = \frac{3n}{2}(2a + (n-1)d)$
Divide both sides by $n$ (assuming $n \ne 0$ ):
$2a + (2n-1)d = \frac{3}{2}(2a + (n-1)d)$
Multiply by 2:
$2(2a + (2n-1)d) = 3(2a + (n-1)d)$
$4a + (4n-2)d = 6a + (3n-3)d$
Rearrange terms to find a relation between $a$ and $d$ :
$(4n-2 - (3n-3))d = 6a - 4a$
$(4n-2-3n+3)d = 2a$
$(n+1)d = 2a$ .

Now we need to find the ratio $S_{3n} : S_n$ :
$\frac{S_{3n}}{S_n} = \frac{\frac{3n}{2}(2a + (3n-1)d)}{\frac{n}{2}(2a + (n-1)d)}$
$\frac{S_{3n}}{S_n} = \frac{3(2a + (3n-1)d)}{2a + (n-1)d}$
Substitute $2a = (n+1)d$ into the expression:
$\frac{S_{3n}}{S_n} = \frac{3((n+1)d + (3n-1)d)}{((n+1)d + (n-1)d)}$
$\frac{S_{3n}}{S_n} = \frac{3d(n+1 + 3n-1)}{d(n+1 + n-1)}$
$\frac{S_{3n}}{S_n} = \frac{3(4n)}{2n}$
$\frac{S_{3n}}{S_n} = \frac{12n}{2n} = 6$ .
Thus, the ratio $S_{3n} : S_n$ is $6:1$ .

The final answer is $\boxed{\text{6:1}}$ "
:::

What's Next?

💡 Continue Your CMI Journey

This chapter provides foundational tools for various advanced topics. Concepts of progressions and means are crucial for understanding and solving problems in inequalities, particularly when applying AM-GM-HM. They also lay groundwork for number theory problems involving sequences and functional equations defined by recurrence relations. Furthermore, the study of sums of series connects to limits and convergence in higher mathematics.

Standard progressions

Standard progressions

Chapter Contents

| Topic |

Part 1: Arithmetic progression

Arithmetic Progression

Overview

Learning Objectives

Core Definition

Standard Forms

Common Difference and Reconstruction

Sum of Terms

Arithmetic Mean

Important Properties

Sign and Growth Cases

Minimal Worked Examples

CMI Strategy

Common Mistakes

Practice Questions

Summary

Part 2: Geometric progression

Geometric Progression

Overview

Learning Objectives

Core Definition

Ratio Recovery

Geometric Mean

Sum of a Finite GP

Infinite GP

Inserted Geometric Means

Important Properties

Behaviour by Ratio

Minimal Worked Examples

CMI Strategy

Common Mistakes

Practice Questions

Summary

Part 3: Harmonic progression

Harmonic Progression

Overview

Learning Objectives

Definition

Core Criterion

Harmonic Mean

General Form of an HP

Useful Relations

HP and AP Connection

Minimal Worked Examples

Insertion of Harmonic Means

Sign and Domain Observations

Common Errors

Strategy for CMI-Type Questions

Practice Questions

Summary

Part 4: Means and their relations

Means and Their Relations

Overview

Learning Objectives

Definitions

Main Relations

AM-GM Inequality

GM-HM Inequality

Combined Inequality

Means in Progression Language

Comparing the Means

Minimal Worked Examples

Equality Condition

Common Errors

Strategy for CMI-Type Questions

Practice Questions

Summary

Chapter Summary

Chapter Review Questions

What's Next?

🎯 Key Points to Remember

Related Topics in Algebra and Functions

Graph-based understanding of functions

Polynomial inequalities

Polynomial equations

Operations on functions