Geometric Mean In GP - Practice Questions & MCQ

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 $G=\sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdot \ldots \ldots \cdot a_n}$.

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

Insertion of $\mathbf{n}$-Geometric Mean Between $\mathbf{a}$ and $\mathbf{b}$
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, $a, \mathrm{G}_1, \mathrm{G}_2, \mathrm{G}_3 \ldots, \mathrm{G}_{\mathrm{n}}, b_{\text {is an }}$. P . Clearly, this $\mathrm{G} . \mathrm{P}$. contains $\mathrm{n}+2$ terms. now, $b=(\mathrm{n}+2)^{\mathrm{th}}$ term $=\operatorname{ar}^{\mathrm{n}+2-1}$

$
\therefore r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}
$

[where, $\mathrm{r}=$ common ratio]

$
\begin{aligned}
& \therefore \mathrm{G}_1=\mathrm{ar}, \mathrm{G}_2=\mathrm{ar}^2, \mathrm{G}_3=\mathrm{ar}^3, \ldots, \mathrm{Ga}_{\mathrm{n}}=\mathrm{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
\end{aligned}
$

$\mathrm{G}_{\mathrm{n}}=\mathrm{a}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{\mathrm{n}}{\mathrm{n}+1}}$

Important Property of GM
The product of $n$ geometric mean between $a$ and $b$ is equal to the $\mathrm{n}^{\text {th }}$ 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, 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}-\mathrm{times}}\right)\left(r^{1+2+3+\ldots+n}\right) \\
& \Rightarrow a^n\left(r^{\left.\left(\frac{n(n+1)}{2}\right)\right)} \quad\left(1+2+\ldots . .+n=\frac{n(n+1)}{2} U \text { sing sum of } A P\right)\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}
$