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# 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) TrueB) False

Last updated date: 20th Mar 2023
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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.

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.