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A). The distribution is symmetric

B). All the values are same

C). The distribution is unimodal

D). None of these

Answer

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Hint: Here, we will use the formulas for AM, GM and HM of two numbers.

Let us suppose two numbers in any series be $a$and $b$

Given, ${\text{AM}} = {\text{GM}} = {\text{HM}}$

As we know that Arithmetic mean of two numbers $a$and $b$ is ${\text{AM}} = \dfrac{{a + b}}{2}$

Geometric mean of two numbers $a$and $b$ is ${\text{GM}} = \sqrt {ab} $

Harmonic mean of two numbers $a$and $b$ is ${\text{HM}} = \dfrac{{2ab}}{{a + b}}$

Now, consider ${\text{AM}} = {\text{GM}} \Rightarrow \dfrac{{a + b}}{2} = \sqrt {ab} $

Squaring above equation both sides we get

\[

\Rightarrow {\left( {\dfrac{{a + b}}{2}} \right)^2} = ab \Rightarrow \dfrac{{{a^2} + {b^2} + 2ab}}{4} = ab \Rightarrow {a^2} + {b^2} + 2ab = 4ab \\

\Rightarrow {a^2} + {b^2} - 2ab = 0 \Rightarrow {\left( {a - b} \right)^2} = 0 \Rightarrow a = b \\

\]

Now, consider $

{\text{AM}} = {\text{HM}} \Rightarrow \dfrac{{a + b}}{2} = \dfrac{{2ab}}{{a + b}} \Rightarrow {\left( {a + b} \right)^2} = 4ab \Rightarrow {a^2} + {b^2} + 2ab = 4ab \\

\Rightarrow {a^2} + {b^2} - 2ab = 0 \Rightarrow {\left( {a - b} \right)^2} = 0 \Rightarrow a = b \\

$

Now, consider ${\text{GM}} = {\text{HM}} \Rightarrow \sqrt {ab} = \dfrac{{2ab}}{{a + b}} \Rightarrow \left( {a + b} \right)\sqrt {ab} = 2ab$

Squaring above equation both sides we get

$

\Rightarrow ab{\left( {a + b} \right)^2} = {\left( {2ab} \right)^2} \Rightarrow {\left( {a + b} \right)^2} = 4ab \Rightarrow {a^2} + {b^2} - 2ab = 0 \\

\Rightarrow {\left( {a - b} \right)^2} = 0 \Rightarrow a = b \\

$

Hence, considering all the possibilities we are always getting that both the numbers in the given series are equal to each other. So, in general we can say that all the values are equal in the series where ${\text{AM}} = {\text{GM}} = {\text{HM}}$.

Therefore, option B is correct.

Note- In these types of problems, we consider any two numbers and apply the formulas for AM, GM and HM in order to find the relation between the assumed numbers.

Let us suppose two numbers in any series be $a$and $b$

Given, ${\text{AM}} = {\text{GM}} = {\text{HM}}$

As we know that Arithmetic mean of two numbers $a$and $b$ is ${\text{AM}} = \dfrac{{a + b}}{2}$

Geometric mean of two numbers $a$and $b$ is ${\text{GM}} = \sqrt {ab} $

Harmonic mean of two numbers $a$and $b$ is ${\text{HM}} = \dfrac{{2ab}}{{a + b}}$

Now, consider ${\text{AM}} = {\text{GM}} \Rightarrow \dfrac{{a + b}}{2} = \sqrt {ab} $

Squaring above equation both sides we get

\[

\Rightarrow {\left( {\dfrac{{a + b}}{2}} \right)^2} = ab \Rightarrow \dfrac{{{a^2} + {b^2} + 2ab}}{4} = ab \Rightarrow {a^2} + {b^2} + 2ab = 4ab \\

\Rightarrow {a^2} + {b^2} - 2ab = 0 \Rightarrow {\left( {a - b} \right)^2} = 0 \Rightarrow a = b \\

\]

Now, consider $

{\text{AM}} = {\text{HM}} \Rightarrow \dfrac{{a + b}}{2} = \dfrac{{2ab}}{{a + b}} \Rightarrow {\left( {a + b} \right)^2} = 4ab \Rightarrow {a^2} + {b^2} + 2ab = 4ab \\

\Rightarrow {a^2} + {b^2} - 2ab = 0 \Rightarrow {\left( {a - b} \right)^2} = 0 \Rightarrow a = b \\

$

Now, consider ${\text{GM}} = {\text{HM}} \Rightarrow \sqrt {ab} = \dfrac{{2ab}}{{a + b}} \Rightarrow \left( {a + b} \right)\sqrt {ab} = 2ab$

Squaring above equation both sides we get

$

\Rightarrow ab{\left( {a + b} \right)^2} = {\left( {2ab} \right)^2} \Rightarrow {\left( {a + b} \right)^2} = 4ab \Rightarrow {a^2} + {b^2} - 2ab = 0 \\

\Rightarrow {\left( {a - b} \right)^2} = 0 \Rightarrow a = b \\

$

Hence, considering all the possibilities we are always getting that both the numbers in the given series are equal to each other. So, in general we can say that all the values are equal in the series where ${\text{AM}} = {\text{GM}} = {\text{HM}}$.

Therefore, option B is correct.

Note- In these types of problems, we consider any two numbers and apply the formulas for AM, GM and HM in order to find the relation between the assumed numbers.