# If one G.M. $G$ and two geometric means $p$ and $q$ be inserted between any two given numbers, then ${{G}^{2}}=\left( 2p-q \right)\left( 2q-p \right)$ .

A) True

B) False

Last updated date: 20th Mar 2023

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Answer

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Hint: The given problem is related to the geometric mean of two numbers. Here we will use the formulae related to the insertion of geometric means between two numbers.

Complete step-by-step answer:

Before proceeding with the solution, first, we will understand the concept of the geometric mean.

The geometric mean of a series with $n$ terms is defined as the ${{n}^{th}}$ root of the product of the terms of the series.

For two numbers, the geometric mean is defined as the square root of the product of the two numbers.

If $n$ geometric means are inserted between two numbers, then the series formed as such will be a geometric progression.

Now, coming to the question, it is given that $G$ is the geometric mean of two numbers. So, let the two numbers be $A$ and $B$. Since $G$ is the geometric mean of $A$ and $B$, so $A$, $G$, and $B$ will be in geometric progression.

So, $\dfrac{G}{A}=\dfrac{B}{G}$.

$\Rightarrow {{G}^{2}}=AB$

Now, it is also given that two geometric means $p$ and $q$ are also inserted between the two given numbers $A$ and $B$.

So, $A,p,q$ and $B$ are in geometric progression.

So, $\dfrac{p}{A}=\dfrac{q}{p}=\dfrac{B}{q}$.

$\Rightarrow {{p}^{2}}=Aq....(i)$ and ${{q}^{2}}=Bp.....(ii)$

From equation$(i)$ , we have ${{p}^{2}}=Aq$.

$\Rightarrow A=\dfrac{{{p}^{2}}}{q}$

From equation$(ii)$ , we have ${{q}^{2}}=Bp$.

$\Rightarrow B=\dfrac{{{q}^{2}}}{p}$

So, $AB=\dfrac{{{p}^{2}}}{q}\times \dfrac{{{q}^{2}}}{p}$.

$=pq$

Now, we have ${{G}^{2}}=AB$ and $AB=pq$.

So, ${{G}^{2}}=pq$.

Hence, the statement that ${{G}^{2}}=\left( 2p-q \right)\left( 2q-p \right)$ is false.

Therefore, the answer is option B.

Note: Don’t get confused between arithmetic mean and geometric mean. The geometric mean of a series with $n$ terms is defined as the ${{n}^{th}}$ root of the product of the terms of the series, whereas the arithmetic mean of a series is defined as the average of the series.

Complete step-by-step answer:

Before proceeding with the solution, first, we will understand the concept of the geometric mean.

The geometric mean of a series with $n$ terms is defined as the ${{n}^{th}}$ root of the product of the terms of the series.

For two numbers, the geometric mean is defined as the square root of the product of the two numbers.

If $n$ geometric means are inserted between two numbers, then the series formed as such will be a geometric progression.

Now, coming to the question, it is given that $G$ is the geometric mean of two numbers. So, let the two numbers be $A$ and $B$. Since $G$ is the geometric mean of $A$ and $B$, so $A$, $G$, and $B$ will be in geometric progression.

So, $\dfrac{G}{A}=\dfrac{B}{G}$.

$\Rightarrow {{G}^{2}}=AB$

Now, it is also given that two geometric means $p$ and $q$ are also inserted between the two given numbers $A$ and $B$.

So, $A,p,q$ and $B$ are in geometric progression.

So, $\dfrac{p}{A}=\dfrac{q}{p}=\dfrac{B}{q}$.

$\Rightarrow {{p}^{2}}=Aq....(i)$ and ${{q}^{2}}=Bp.....(ii)$

From equation$(i)$ , we have ${{p}^{2}}=Aq$.

$\Rightarrow A=\dfrac{{{p}^{2}}}{q}$

From equation$(ii)$ , we have ${{q}^{2}}=Bp$.

$\Rightarrow B=\dfrac{{{q}^{2}}}{p}$

So, $AB=\dfrac{{{p}^{2}}}{q}\times \dfrac{{{q}^{2}}}{p}$.

$=pq$

Now, we have ${{G}^{2}}=AB$ and $AB=pq$.

So, ${{G}^{2}}=pq$.

Hence, the statement that ${{G}^{2}}=\left( 2p-q \right)\left( 2q-p \right)$ is false.

Therefore, the answer is option B.

Note: Don’t get confused between arithmetic mean and geometric mean. The geometric mean of a series with $n$ terms is defined as the ${{n}^{th}}$ root of the product of the terms of the series, whereas the arithmetic mean of a series is defined as the average of the series.

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