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Question: The AM, GM and HM in any series are equal then A). The distribution is symmetric B). All the val...

The AM, GM and HM in any series are equal then
A). The distribution is symmetric
B). All the values are same
C). The distribution is unimodal
D). None of these

Explanation

Solution

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

Let us suppose two numbers in any series be aaand bb
Given, AM=GM=HM{\text{AM}} = {\text{GM}} = {\text{HM}}
As we know that Arithmetic mean of two numbers aaand bb is AM=a+b2{\text{AM}} = \dfrac{{a + b}}{2}
Geometric mean of two numbers aaand bb is GM=ab{\text{GM}} = \sqrt {ab}
Harmonic mean of two numbers aaand bb is HM=2aba+b{\text{HM}} = \dfrac{{2ab}}{{a + b}}
Now, consider AM=GM⇒a+b2=ab{\text{AM}} = {\text{GM}} \Rightarrow \dfrac{{a + b}}{2} = \sqrt {ab}
Squaring above equation both sides we get

⇒(a+b2)2=ab⇒a2+b2+2ab4=ab⇒a2+b2+2ab=4ab ⇒a2+b2−2ab=0⇒(a−b)2=0⇒a=b  \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 AM=HM⇒a+b2=2aba+b⇒(a+b)2=4ab⇒a2+b2+2ab=4ab ⇒a2+b2−2ab=0⇒(a−b)2=0⇒a=b  {\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 GM=HM⇒ab=2aba+b⇒(a+b)ab=2ab{\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
⇒ab(a+b)2=(2ab)2⇒(a+b)2=4ab⇒a2+b2−2ab=0 ⇒(a−b)2=0⇒a=b  \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 AM=GM=HM{\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.