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

If the harmonic between a and b be H, then the value of 1H−a+1H−b\dfrac{1}{{H - a}} + \dfrac{1}{{H - b}}
A) a+ba + b
B) abab
C) 1a+1b\dfrac{1}{a} + \dfrac{1}{b}
D) 1a−1b\dfrac{1}{a} - \dfrac{1}{b}

Explanation

Solution

According to given in the question we have to determine the value of 1H−a+1H−b\dfrac{1}{{H - a}} + \dfrac{1}{{H - b}} if the harmonic between a and b be H. So, first of all we have to understand about harmonic mean as mentioned below:
Harmonic mean: The harmonic mean actually a type of numerical average and it is calculated by dividing the number of observations by finding the reciprocal of each number in the given sequence or series hence, the harmonic mean is the reciprocal of the arithmetic mean of the reciprocal and the value of harmonic mean between a and b is given below:

Formula used: H=2aba+b................(A) H = \dfrac{{2ab}}{{a + b}}................(A)
Now, we have to substitute the value of H in the given expression 1H−a+1H−b\dfrac{1}{{H - a}} + \dfrac{1}{{H - b}} after that we have to solve the obtained expression we have to take the L.C.M and after that we have to take the common terms that can be and eliminate the terms that can be to obtain the value of the given expression.

Complete step-by-step answer:
Step 1: First of all we have to substitute the value of H as from the formula (A) which is mentioned in the solution hint. Hence,
=12aba+b−a+12aba+b−b...............(1)= \dfrac{1}{{\dfrac{{2ab}}{{a + b}} - a}} + \dfrac{1}{{\dfrac{{2ab}}{{a + b}} - b}}...............(1)
Step 2: Now, to solve the expression (1) we have to find the L.C.M of the term obtained,
=12ab−a(a+b)a+b+12ab−b(a+b)a+b =12ab−a2−aba+b+12ab−ab−b2a+b = \dfrac{1}{{\dfrac{{2ab - a(a + b)}}{{a + b}}}} + \dfrac{1}{{\dfrac{{2ab - b(a + b)}}{{a + b}}}} \\\ = \dfrac{1}{{\dfrac{{2ab - {a^2} - ab}}{{a + b}}}} + \dfrac{1}{{\dfrac{{2ab - ab - {b^2}}}{{a + b}}}}
Now, we have to subtract the terms of the expression as obtained just above and on rearranging the terms,
=1ab−a2a+b+1ab−b2a+b =a+bab−a2+a+bab−b2............(2) = \dfrac{1}{{\dfrac{{ab - {a^2}}}{{a + b}}}} + \dfrac{1}{{\dfrac{{ab - {b^2}}}{{a + b}}}} \\\ = \dfrac{{a + b}}{{ab - {a^2}}} + \dfrac{{a + b}}{{ab - {b^2}}}............(2)
Step 3: Now, we have to take the terms common that can be as obtained in the expression (2). Hence,
=a+bb−a(1a−1b) =a+bb−a×b−aab = \dfrac{{a + b}}{{b - a}}\left( {\dfrac{1}{a} - \dfrac{1}{b}} \right) \\\ = \dfrac{{a + b}}{{b - a}} \times \dfrac{{b - a}}{{ab}}
Now, on eliminating the terms (b-a) as obtained in the expression just above,
=a+bab =1a+1b = \dfrac{{a + b}}{{ab}} \\\ = \dfrac{1}{a} + \dfrac{1}{b}
Final solution: Hence, with the help of formula (A) we have obtained the value of the given expression1H−a+1H−b=1a+1b\dfrac{1}{{H - a}} + \dfrac{1}{{H - b}} = \dfrac{1}{a} + \dfrac{1}{b}.

Therefore option (C) is the correct answer.

Note: The simple way to determine the harmonic mean that it is the reciprocal of the given arithmetic mean of the reciprocals of the observations.
It is used to calculate the average of the group of the numbers and the number of elements will be averaged and divided by the sum of reciprocals of the elements.