Relation Between AM, GM and HM and Formula

Arithmetic Mean, Geometric Mean, and Harmonic Mean are three different types of measures of central tendency. If the terms of a sequence follow some pattern that can be defined by an explicit formula in n, then the sequence is called a progression. We have three types of progression Arithmetic Progression, Geometric progression, and harmonic progression. In real life, we use this Progression in electrical gadgets, machines, or the generation of power. It is also used to calculate the degree to which water boils when its temperature increases by the same amount.

This Story also Contains

1. Arithmetic Progression
2. Geometric Progression
3. Harmonic Progression
4. Application of A.M., G.M., and H.M.
5. Solved Examples Based on Relation Between AM, GM, and HM

Arithmetic Progression

An arithmetic progression is a sequence in which each term increases or decreases by a constant term or fixed number. This fixed number is called the common difference of an AP and is generally denoted by ' $d$ '.
$\mathrm{Eg}, 1,4,7,10, \ldots$. is an AP with a common difference 3
Also, $2,1,0,-1, \ldots$ is an AP with a common difference of -1
In AP, the first term is generally denoted by 'a.
If three terms are in AP , then the middle term is called the Arithmetic Mean (A.M.) of the other two numbers. So if $a, b$, and $c$ are in A.P., then $b$ is $A M$ of a and c .

If $a_1, a_2, a_3, \ldots ., a_n$ are n positive numbers, then the Arithmetic Mean of these numbers is given by

$
A=\frac{a_1+a_2+a_3+\ldots . .+a_n}{n}
$

Geometric Progression

A geometric progression or geometric sequence is a sequence where the first term is non-zero and the ratio between consecutive terms is always constant. The 'constant factor' is called the common ratio and is denoted by ' $r$ '. $r$ is also a non-zero number.

The first term of a G.P. is usually denoted by 'a'.
If each term of a G.P. is multiplied by a fixed constant or divided by a nonzero fixed constant then the resulting series is also in G.P. with the same common ratio as the original series.

If each term of a G.P. is raised to some real number $m$, then the resulting series is also in G.P.

If three terms are in G.P., then the middle term is called the Geometric Mean (G.M.) of the other two numbers. So if $a, b$, and $c$ are in G.P., then $b$ is GM of a and $c$,

If $a_1, a_2, a_3, \ldots ., a_n$ are $n$ positive numbers, then the Geometric Mean of these numbers is given by $G=\sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdot \ldots \cdot \cdot a_n}$.

If a and b are two numbers and G is the GM of a and b . Then, $\mathrm{a}, \mathrm{G}$, and b are in geometric progression.

Hence, $G=\sqrt{a \cdot b}$

Harmonic Progression

A Harmonic Progression $(\mathrm{HP})$ is defined as a sequence of real numbers obtained by taking the reciprocals of an Arithmetic Progression that excludes 0 .

A sequence $a_1, a_2, a_3, \ldots, a_n, \ldots$ of non-zero numbers is called a harmonic progression if the sequence $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots ., \frac{1}{a_n}, \ldots$. is an arithmetic progression.

OR
Reciprocals of arithmetic progression is a Harmonic progression.
E.g. $\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots \ldots$
form an A.P.
The mean of the given Harmonic Progression is called the Harmonic Mean.

Application of A.M., G.M., and H.M.

Let $A$, $G$, and H be arithmetic, geometric, and harmonic means of two positive real numbers a and b .

Then,

$
\mathrm{A}=\frac{a+b}{2}, \quad \mathrm{G}=\sqrt{a \cdot b} \text { and } \mathrm{H}=\frac{2 a b}{a+b}
$

$\begin{equation}
\begin{aligned}
&\begin{aligned}
& \text { 1. } \mathrm{A} \geq \mathrm{G} \geq \mathrm{H} \\
& \mathrm{A}-\mathrm{G}=\frac{a+b}{2}-\sqrt{a b}=\frac{(\sqrt{a}-\sqrt{b})^2}{2} \geq 0 \\
& \Rightarrow \mathrm{A}-\mathrm{G} \geq 0 \\
& \Rightarrow \mathrm{A} \geq \mathrm{G}
\end{aligned}\\
&\text { ........(i) }
\end{aligned}
\end{equation}$

Note that A = G when a = b

Note that $\mathrm{A}=\mathrm{G}$ when $\mathrm{a}=\mathrm{b}$
Now,

$
\begin{aligned}
& \mathrm{G}-\mathrm{H}=\sqrt{a b}-\frac{2 a b}{a+b} \\
&=\sqrt{a b}\left(\frac{a+b-2 \sqrt{a b}}{a+b}\right) \\
&=\frac{\sqrt{a b}}{a+b}(\sqrt{a}-\sqrt{b})^2 \geq 0 \\
& \Rightarrow \mathrm{G} \geq \mathrm{H}
\end{aligned}
$

Again $\mathrm{G}=\mathrm{H}$ when $\mathrm{a}=\mathrm{b}$

From (i) and (ii) we get

$
\mathrm{A} \geq \mathrm{G} \geq \mathrm{H}
$

Note :
- when $\mathrm{a}=\mathrm{b}$ then only, $\mathrm{A}=\mathrm{G}=\mathrm{H}$
- The same relation $A \geq G \geq H$ can be applied for AM, GM and HM of more than 2 positive real numbers

If $a_1, a_2, a_3, \ldots \ldots, a_n$ are $n$ positive real numbers, then
A $=$ A.M. of $a_1, a_2, a_3, \ldots ., a_n=\frac{a_1+a_2+a_3+\ldots .+a_n}{n}$
G $=$ G.M. of $a_1, a_2, a_3, \ldots ., a_n=\left(a_1 \cdot a_2 \cdot a_3 \ldots \ldots . a_n\right)^{\frac{1}{n}}$
$\mathrm{H}=$ H.M. of $a_1, a_2, a_3, \ldots \ldots, a_n=\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\ldots \ldots+\frac{1}{a_n}}$
In such case also $A \geq G \geq H$
And $\mathrm{A}=\mathrm{G}=\mathrm{H}$, when $a_1=a_2=a_3=\ldots \ldots=a_n$

2. $A, G$ and $H$ of 2 positive real numbers form a geometric progression, i.e. $\mathbf{G}^{\mathbf{2}}=\mathbf{A H}$.

we have,

$
\begin{aligned}
\mathrm{A} \cdot \mathrm{H} & =\frac{a+b}{2} \times \frac{2 a b}{a+b} \\
& =a b=(\sqrt{a b})^2=\mathrm{G}^2
\end{aligned}
$

Hence, $\quad \mathrm{G}^2=\mathrm{AH}$

Recommended Video Based on Relation Between AP, GP and HP:

Solved Examples Based on Relation Between AM, GM, and HM

Example 1: If $\mathrm{AM}=\mathrm{GM}=\mathrm{HM}$ for two positive numbers a and b then which of the following is NEVER true?
1) $a=b$
2) $a=1$
3) $a=10$
4) $a>b$

Solution

As we learned

Relation between $\mathrm{AM}, \mathrm{GM}$, and HM of two positive numbers

$
\begin{aligned}
& \quad A M=G M=H M \\
& \text { if } a=b
\end{aligned}
$

If both numbers are equal, all the three means are identical.
In this Question,
Both $a$ and $b$ should be equal
Option 2 and 3 are possible as $\mathrm{a}=\mathrm{b}=1$ or $\mathrm{a}=\mathrm{b}=10$ will give $\mathrm{AM}=\mathrm{GM}=\mathrm{HM}$
Hence, the answer is the option (4).

Example 2: Let $x, y, z$ be positive real numbers such that $x+y+z=12$ and $x^3 y^4 z^5=(0.1)(600)^3$. Then $x^3+y^3+z^3$ is equal to :
1) 270
2) 258
3) 342
4) 216

Solution

This is the AM between these 12 numbers

As we have learned
Relation between $\mathrm{AM}, \mathrm{GM}$, and HM of two positive numbers -

$
A M \geqslant G M \geqslant H M
$

Now,

$
\begin{aligned}
& x+y+z=12 \\
& \left(\frac{x}{3}+\frac{x}{3}+\frac{x}{3}\right)+\left(\frac{y}{4}+\frac{y}{4}+\frac{y}{4}+\frac{y}{4}\right)+\left(\frac{z}{5}+\frac{z}{5}+\frac{z}{5}+\frac{z}{5}+\frac{z}{5}\right)=12 \\
& \frac{\left(\frac{x}{3}+\frac{x}{3}+\frac{x}{3}\right)+\left(\frac{y}{4}+\frac{y}{4}+\frac{y}{4}+\frac{y}{4}\right)+\left(\frac{z}{5}+\frac{z}{5}+\frac{z}{5}+\frac{z}{5}+\frac{z}{5}\right)}{12}=1
\end{aligned}
$

This is the AM between these 12 numbers

Now let us find GM between these 12 numbers

$
\begin{aligned}
& \text { G.M }=\left[(x / 3)^3(y / 4)^4(z / 5)^5\right]^{1 / 12} \\
& \text { G.M }=\left[\frac{(x)^3(y)^4(z)^5}{(3)^3(4)^4(5)^5}\right]^{1 / 12}
\end{aligned}
$

Now, as $A . M \geq$ G.M

$
\begin{aligned}
& 1 \geq\left[\frac{(x)^3(y)^4(z)^5}{(3)^3(4)^4(5)^5}\right]^{1 / 12} \\
& \Rightarrow\left(x^3 y^4 z^5\right) \leq(0.1)(600)^3
\end{aligned}
$

But given $x^3 y^4 z^5=(0.1)(600)^3$
Which means as given A.M $=$ G.M

$
\Rightarrow x / 3=y / 4=z / 5
$

From above result and $x+y+z=12$, we get $x=3, y=4, z=5$

$
\Rightarrow x^3+y^3+z^3=3^3+4^3+5^3=216
$
Hence, the answer is the option (4).

Example 3: If AM and HM of two positive numbers are 125 and 5 respectively, then their GM is:
1) 5
2) 125
3) 25
4) 1

Solution

As we learned

Relation between $\mathrm{AM}, \mathrm{GM}$, and HM of two positive numbers

$
(G M)^2=(A M)(H M)
$

In this Question,

$
(G M)^2=5 \times 125=625 \Rightarrow G M=25
$

Hence, the answer is the option (3).
Example 4: Let $\mathrm{x}, \mathrm{y}$ be positive real numbers and $\mathrm{m}, \mathrm{n}$ positive integers. $\frac{x^m y^n}{\left(1+x^{2 m}\right)\left(1+y^{2 n}\right)}$ is
1) 1
2) 0.5
3) 0.25
4) 1.5

Solution

Relation between $\mathrm{AM}, \mathrm{GM}$, and HM of two positive numbers -

$
A M \geqslant G M \geqslant H M
$

Now,

$
\begin{aligned}
& \frac{x^m y^n}{\left(1+x^{2 m}\right)\left(1+y^{2 n}\right)} \\
& =\frac{1}{\left(x^m+\frac{1}{x^m}\right)\left(y^n+\frac{1}{y^n}\right)} \\
& A \cdot M \cdot \geq G \cdot M \\
& \frac{\left(x^m+\frac{1}{x^m}\right)}{2} \geq \sqrt{\left(x^m \cdot \frac{1}{x^m}\right)} \text { and } \frac{\left(y^m+\frac{1}{y^m}\right)}{2} \geq \sqrt{\left(y^m \cdot \frac{1}{y^m}\right)} \\
& \left(x^m+\frac{1}{x^m}\right) \geq 2 \text { and }\left(y^m+\frac{1}{y^m}\right) \geq 2
\end{aligned}
$
Multiplying both

$
\left(x^m+\frac{1}{x^m}\right) \cdot\left(y^m+\frac{1}{y^m}\right) \geq 4
$

So, $\frac{1}{\left(x^m+\frac{1}{x^m}\right)\left(y^n+\frac{1}{y^n}\right)}=\frac{x^m y^n}{\left(1+x^{2 m}\right)\left(1+y^{2 n}\right)} \leq \frac{1}{4}$
Hence, the answer is the option (3).
Example 5: If

$
f(x)=\frac{\left(\tan 1^{\circ}\right) x+\log _e(123)}{x \log _e(1234)-\left(\tan 1^{\circ}\right)}, x>0
$

then the least value of
1) 2
2) 4
3) 8
4) 0

Solution

Hence, the answer is the option 2.

$\begin{equation}
\begin{aligned}
&\begin{aligned}
& f(x)=\frac{(\tan 1) x+\log _e 123}{x \log 1234-(\tan 1)} \\
& \text { Let } A=\tan 1, B=\log 123, C=\text { LOG } 1234 \\
& f(x)=\frac{A x+B}{x C-A} \\
& f(f(x))=\frac{A\left(\frac{A x+B}{x C-A}\right)+B}{C\left(\frac{A x+B}{C X-A}\right)-A} \\
& =\frac{A^2 x+A B+x B C-A B}{A C x+B C-A C x+A^2} \\
& =\frac{x\left(A^2+B C\right)}{\left(A^2+B C\right)}=x \\
& f(f(x))=x \\
& f\left(f\left(\frac{4}{x}\right)\right)=\frac{4}{x} \\
& f(f(x))+f\left(f\left(\frac{4}{x}\right)\right) \\
& A M \geq G M \\
& x+\frac{4}{x} \geq 4
\end{aligned}\\
&\text { Hence, the answer is the option } 2 \text {. }
\end{aligned}
\end{equation}$