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
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.
Last updated date: 21st Sep 2023
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