Geometric Mean In GP - Definition and Properties

Geometric Sequence

A 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 $a_1, a_2, a_3 \ldots . a_{n-1}, a_n$ is in geometric progression then, $r=\frac{a_2}{a_1}=\frac{a_3}{a_2}=\ldots=\frac{a_n}{a_{n-1}}$

Eg ,
- $2,6,18,54, \ldots .(a=2, r=3)$
- $4,2,1,1 / 2,1 / 4, \ldots .(a=4, r=1 / 2)$
- $-5,5,-5,5, \ldots \ldots(a=-5, r=-1)$

General Term of a GP

$
\begin{aligned}
& a_1=a=a r^{1-1}\left(1^{\text {st }} \text { term }\right) \\
& a_2=a r=a r^{2-1}\left(2^{\text {nd }} \text { term }\right) \\
& a_3=a r^2=a r^{3-1}\left(3^{\text {rd }} \text { term }\right) \\
& \cdots \\
& \cdots \\
& a_n=a r^{n-1}\left(\mathrm{n}^{\text {th }} \text { term }\right)
\end{aligned}
$
So, the general term or $\mathrm{n}^{\text {th }}$ term of a geometric progression is $a_n=a r^{n-1}$

Geometric Mean

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, $G=\sqrt{a \cdot b}$
Insertion of n-Geometric Mean Between a and b
Let $r$ be the common ratio of this GP.
now, $\mathrm{b}=(\mathrm{n}+2)^{\text {th }}$ term $=\mathrm{ar}^{\mathrm{n}+2-1}$
$\therefore r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$
[where, $\mathrm{r}=$ common ratio]
$\therefore \mathrm{G}_1=$ ar, $\mathrm{G}_2=\operatorname{ar}^2, \mathrm{G}_3=\operatorname{ar}^3, \ldots ., \mathrm{Ga}_{\mathrm{n}}=\operatorname{ar}^{\mathrm{n}}$
$\Rightarrow \mathrm{G}_1=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{1}{\mathrm{n}+1}}, \mathrm{G}_2=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{2}{\mathrm{n}+1}}, \mathrm{G}_3=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{3}{\mathrm{n}+1}} \ldots \ldots$
$\mathrm{G}_{\mathrm{n}}=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{\mathrm{n}}{\mathrm{n}+1}}$

Important Property of GM

The product of $n$ geometric mean between $a$ and $b$ is equal to the $n^{t h}$ power of a single geometric mean between $a$ and $b$.
If a and b are two numbers and $\mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3 \ldots, \mathrm{G}_{\mathrm{n}}$ are n -geometric mean between a and b , then $a, \mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3 \ldots, \mathrm{G}_{\mathrm{n}}, b$ will be in geometric progression.
So, the Product of n-G.M's between a and b is

$
\begin{aligned}
& \mathrm{G}_1 \cdot \mathrm{G}_2 \cdot \mathrm{G}_3 \cdot \ldots \cdot \mathrm{G}_{\mathrm{n}}=(a r)\left(a r^2\right)\left(a r^3\right) \ldots\left(a r^n\right) \\
& \Rightarrow\left(a^{1+1+1+\ldots \mathrm{n}-\text { times }}\right)\left(r^{1+2+3+\ldots+n}\right) \\
& \Rightarrow a^n\left(r^{\left(\frac{n(n+1)}{2}\right)}\right) \quad\left(1+2+\ldots . \cdot+n=\frac{n(n+1)}{2} U \text { sing sum of } A P\right) \\
& \text { replace } r \text { with }\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\left(\frac{1}{\mathrm{n}+1}\right)} \\
& \Rightarrow a^n \cdot\left[\left(\frac{b}{a}\right)^{\frac{1}{n+1}}\right]^{\frac{n(n+1)}{2}}=a^n\left(\frac{b}{a}\right)^{\frac{n}{2}} \\
& \Rightarrow(a)^{\frac{n}{2}}(b)^{\frac{n}{2}}=(\sqrt{a \cdot b})^n \\
& =[\mathrm{G} \cdot \mathrm{M} . \text { of } a \text { and } b]^{\mathrm{n}}
\end{aligned}
$

Recommended Video Based on Geometric Mean

Solved Examples Based on Geometric Mean

Example 1: Consider two G.Ps. $2,2^2, 2^3, \ldots$ and $4,4^2, 4^3, \ldots$ of 60 and ' $n$ ' terms respectively. If the geometric mean of all the $60+n$ terms is $\qquad$ is equal to:

[JEE
MAINS 2022]
Solution: Given: $\left(\left(2^1 \cdot 2^2 \cdots \cdots 2^{60}\right)\left(4^1 \cdot 4^2 \cdots 4^{\mathrm{n}}\right)\right)^{\frac{1}{60+n}}=2^{\frac{225}{8}}$

$
\begin{aligned}
& \Rightarrow\left(2^{30 \times 61} \cdot 4^{\frac{\mathrm{n}(\mathrm{n}+1)}{2}}\right)^{\frac{1}{60+n}}=2^{\frac{225}{8}} \\
& \Rightarrow 2^{1830+\mathrm{n}^2+\mathrm{n}}=2^{\frac{(225)(60+\mathrm{n})}{8}} \\
& \Rightarrow 8 \mathrm{n}^2-217 \mathrm{n}+1140=0 \\
& \Rightarrow \mathrm{n}=20, \frac{57}{8}
\end{aligned}
$

Now,

$
\begin{aligned}
\sum_{k=1}^n n k-k^2 & =\frac{n^2(n+1)}{2}-\frac{n(n+1)(2 n+1)}{6} \\
& =1330
\end{aligned}
$

Hence, the answer is 1330

Example 2: If $m$ arithmetic means $(A . M s)$ and three geometric means $(G . M s)$ are inserted between 3 and 243 such that $4^{\text {th }}$ A.M. is equal to $2^{\text {nd }}$ G.M., then $m$ is equal to $\qquad$
[JEE MAINS 2020]

Solution: Let $\mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3 \ldots, \mathrm{A}_{\mathrm{n}}$ be n arithmetic mean between two numbers a and b . Then, $a, \mathrm{~A}_1, \mathrm{~A}_2, \mathrm{~A}_3 \ldots, \mathrm{A}_{\mathrm{n}}, b$ is an A.P. Let d be the common difference of this A.P. Clearly, this A.P. contains $n+2$ terms, and

Let $\mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3 \ldots, \mathrm{G}_{\mathrm{n}}$ be n geometric mean between two numbers a and b . Then, $a, \mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3 \ldots, \mathrm{G}_{\mathrm{n}}, b$ is a G.P. Clearly, this G.P. contains $\mathrm{n}+2$ terms.

Given,
$3, \mathrm{~A}_1, \mathrm{~A}_2 \ldots \ldots . \mathrm{A}_{\mathrm{m}}, 243 \Rightarrow \mathrm{d}=\frac{243-3}{\mathrm{~m}+1}=\frac{240}{\mathrm{~m}+1}$ Now $3, \mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3, 243$

$
\begin{aligned}
& \Rightarrow \mathrm{r}=\left(\frac{243}{3}\right)^{\frac{1}{3+1}}=3 \\
\therefore \quad & \mathrm{A}_4=\mathrm{G}_2 \\
\Rightarrow \quad & \mathrm{a}+4 \mathrm{~d}=\mathrm{ar}^2 \\
\Rightarrow & 3+4\left(\frac{240}{\mathrm{~m}+1}\right)=3(3)^2 \\
& \Rightarrow \mathrm{m}=39
\end{aligned}
$

Hence, the answer is 39

Example 3: If the arithmetic mean of two numbers a and $b, a>b>0$, is

$
\frac{a+b}{a-b}
$

is equal to:
five times their geometric mean, then $a$

$
A=\frac{a+b}{2}
$

and

The geometric mean of two numbers (GM) $-G M=\sqrt{a b}$
In this Question,

$
\begin{aligned}
& \frac{a+b}{2}=5 \sqrt{a b} \\
& a+b=10 \sqrt{a b}
\end{aligned}
$

We know,

$
\begin{aligned}
& (a-b)^2=(a+b)^2-4 a b \\
& =(10 \sqrt{a b})^2-4 a b=100 a b-4 a b=96 a b \\
& a-b=4 \sqrt{6} \sqrt{a b}
\end{aligned}
$
Therefore:

$
\frac{a+b}{a-b}=\frac{10 \sqrt{a b}}{4 \sqrt{6} \sqrt{a b}}=5 \frac{\sqrt{6}}{12}
$

Hence, the answer is $\frac{5 \sqrt{6}}{12}$

Example 4: Let G be the geometric mean of two positive numbers a and b , and M be the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ if $\frac{1}{M}: G$ is $4: 5$ then a:b can be:

Solution: We know that the arithmetic mean of two numbers (AM) is given by $A=\frac{a+b}{2}$
and

The geometric mean of two numbers (GM) is given by $G M=\sqrt{a b}$
Now, given $\mathrm{G}^2=\mathrm{ab}$

$
\begin{aligned}
& 2 M=\frac{1}{a}+\frac{1}{b} \\
& \text { and } \frac{\frac{1}{M}}{G}=\frac{4}{5} \\
& \therefore G M=\frac{5}{4}
\end{aligned}
$

Squaring both sides we get,

$
\therefore G^2 M^2=\frac{25}{16}
$

Putting values of G and M ,

$
\begin{aligned}
& a b \times\left(\frac{a+b}{2 a b}\right)^2=\frac{25}{16} \\
& \frac{a b \times(a+b)^2}{4 a^2 b^2}=\frac{25}{16}
\end{aligned}
$

solving equation we get,

$
\therefore 4 a^2+4 b^2-17 a b=0
$

dividing the equation by $b$

$
\therefore 4\left(\frac{a}{b}\right)^2-17\left(\frac{a}{b}\right)+4=0
$

So, we get

$
\therefore \frac{a}{b}=1: 4
$

Hence, the answer is 1:4

Example 5: If $\alpha, \beta$ and $\gamma$ are three consecutive terms of a nonconstant G.P. such that the equations $\alpha x^2+2 \beta x+\gamma=0$ and $x^2+x-1=0$ have a common root, then $\alpha(\beta+\gamma)$ is equal to:

Solution: The geometric mean of two numbers (GM) is given by $G M=\sqrt{a b}$

Condition for Real and equal roots of Quadratic Equation -

$
D=b^2-4 a c=0
$

Now,
$\alpha, \beta$ and $\gamma$ are in G.P. $\Rightarrow \beta^2=\alpha \gamma$
For equation, $\alpha x^2+2 \beta x+\gamma=0$

$
\Delta=4 \beta^2-4 \alpha \gamma=0
$

Hence, roots are equal \& equal to $-\frac{\beta}{\alpha}$ (Using the sum of roots)

$
-\frac{\beta}{\alpha}=-\frac{\gamma}{\beta}(\text { Using GP relation) }
$

Since the given equation has common roots, hence $-\frac{\gamma}{\beta}$ must be the root of $x^2+x-1=0$

$
\begin{aligned}
& =>\frac{\gamma^2}{\beta^2}-\frac{\gamma}{\beta}-1=0 \\
& =>\gamma^2-\gamma \beta-\beta^2=0 \\
& =>\gamma^2=\beta(\gamma+\beta) \\
& =>\gamma \cdot \frac{\beta^2}{\alpha}=\beta(\gamma+\beta) \\
& =>\gamma \cdot \beta=\alpha(\gamma+\beta)
\end{aligned}
$

Hence, the answer is $\beta \gamma$