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Question: If H be the harmonic mean between \(a\) and \(b\) , then the value of \(\dfrac{1}{{(H - a)}} + \dfra...

If H be the harmonic mean between aa and bb , then the value of 1(H−a)+1(H−b)\dfrac{1}{{(H - a)}} + \dfrac{1}{{(H - b)}} is
1)a+b1)a + b
2)ab2)ab
3)1a+1b3)\dfrac{1}{a} + \dfrac{1}{b}
4)1a−1b4)\dfrac{1}{a} - \dfrac{1}{b}

Explanation

Solution

First, we need to know about the concepts of harmonic mean.
The harmonic mean is defined as one of the types of determining the average. It is computed by dividing the number of values in the sequence by the sum of reciprocals of the terms.
Also, which can be obtained by the relation between the Arithmetic mean.
Compare the relationship between these AM, GM, HM, and we get to generalize the common formula for the AM, GM, and HM. Which is used in the question like all three mean required values.
Formula used:
The harmonic mean can be represented as HM=2aba+bHM = \dfrac{{2ab}}{{a + b}}.

Complete step-by-step solution:
Since from the given that we have, H be the harmonic mean between aa and bb.
By the Harmonic formula, we have, H=2aba+bH = \dfrac{{2ab}}{{a + b}} (where H is the harmonic mean)
Now subtract the value aa on both sides of the equation, we get H−a=2aba+b−aH - a = \dfrac{{2ab}}{{a + b}} - a
Further solving using the cross product, we get H−a=2aba+b−a⇒2ab−a(a+b)a+bH - a = \dfrac{{2ab}}{{a + b}} - a \Rightarrow \dfrac{{2ab - a(a + b)}}{{a + b}}
Now by the multiplication and subtraction operation, we have H−a=2ab−a(a+b)a+b⇒2ab−a2−aba+b⇒ab−a2a+bH - a = \dfrac{{2ab - a(a + b)}}{{a + b}} \Rightarrow \dfrac{{2ab - {a^2} - ab}}{{a + b}} \Rightarrow \dfrac{{ab - {a^2}}}{{a + b}}
Taking reciprocal, we get H−a=ab−a2a+b⇒1H−a=a+bab−a2H - a = \dfrac{{ab - {a^2}}}{{a + b}} \Rightarrow \dfrac{1}{{H - a}} = \dfrac{{a + b}}{{ab - {a^2}}}
Similarly, subtract the value bbon both sides of the equation, we get H−b=2aba+b−bH - b = \dfrac{{2ab}}{{a + b}} - b
Further solving using the cross product, we get H−b=2aba+b−b⇒2ab−b(a+b)a+bH - b = \dfrac{{2ab}}{{a + b}} - b \Rightarrow \dfrac{{2ab - b(a + b)}}{{a + b}}
Now by the multiplication and subtraction operation, we have H−b=2ab−b(a+b)a+b⇒2ab−b2−aba+b⇒ab−b2a+bH - b = \dfrac{{2ab - b(a + b)}}{{a + b}} \Rightarrow \dfrac{{2ab - {b^2} - ab}}{{a + b}} \Rightarrow \dfrac{{ab - {b^2}}}{{a + b}}
Taking reciprocal, we get H−b=ab−b2a+b⇒1H−b=a+bab−b2H - b = \dfrac{{ab - {b^2}}}{{a + b}} \Rightarrow \dfrac{1}{{H - b}} = \dfrac{{a + b}}{{ab - {b^2}}}
Add both values we get 1H−a+1H−b=a+bab−a2+a+bab−b2\dfrac{1}{{H - a}} + \dfrac{1}{{H - b}} = \dfrac{{a + b}}{{ab - {a^2}}} + \dfrac{{a + b}}{{ab - {b^2}}}
Further solving we get, 1H−a+1H−b=a+ba(b−a)+a+bb(a−b)⇒a+b(b−a)[1a−1b]\dfrac{1}{{H - a}} + \dfrac{1}{{H - b}} = \dfrac{{a + b}}{{a(b - a)}} + \dfrac{{a + b}}{{b(a - b)}} \Rightarrow \dfrac{{a + b}}{{(b - a)}}[\dfrac{1}{a} - \dfrac{1}{b}] (by taking out the common terms)
By the cross product we have, 1H−a+1H−b=a+b(b−a)[1a−1b]⇒a+b(b−a)[b−aab]⇒a+bab\dfrac{1}{{H - a}} + \dfrac{1}{{H - b}} = \dfrac{{a + b}}{{(b - a)}}[\dfrac{1}{a} - \dfrac{1}{b}] \Rightarrow \dfrac{{a + b}}{{(b - a)}}[\dfrac{{b - a}}{{ab}}] \Rightarrow \dfrac{{a + b}}{{ab}}
Hence, we get 1H−a+1H−b=a+bab⇒1a+1b\dfrac{1}{{H - a}} + \dfrac{1}{{H - b}} = \dfrac{{a + b}}{{ab}} \Rightarrow \dfrac{1}{a} + \dfrac{1}{b} where a+bab=1a+1b\dfrac{{a + b}}{{ab}} = \dfrac{1}{a} + \dfrac{1}{b}
Therefore option 3)1a+1b3)\dfrac{1}{a} + \dfrac{1}{b} is correct.

Note: AM is the average or mean of the given set of numbers which is computed by adding all the terms in the set of numbers and dividing the sum by the given total number of terms.
Thus, we get AM=a+b2AM = \dfrac{{a + b}}{2}where a and b are the sum of the terms and two is the total count.
The geometric mean is the mean value or the central term in the set of numbers in the geometric progression. Geometric means of sequence with the n terms is computed as the nth root of the product of all the terms in the sequence taken.
Thus, we get GM=abGM = \sqrt {ab}
Now apply AM=a+b2AM = \dfrac{{a + b}}{2} and its reciprocal is the HM.
Thus, we get HM=21a+1b⇒2aba+bHM = \dfrac{2}{{\dfrac{1}{a} + \dfrac{1}{b}}} \Rightarrow \dfrac{{2ab}}{{a + b}}