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Question: \({H_1},{H_2}\) are 2 H.M.s between \(a\) and \(b\) , then \[\dfrac{{{H_1} + {H_2}}}{{{H_1}.{H_2}}} ...

H1,H2{H_1},{H_2} are 2 H.M.s between aa and bb , then H1+H2H1.H2=\dfrac{{{H_1} + {H_2}}}{{{H_1}.{H_2}}} =
A) aba+b\dfrac{{ab}}{{a + b}}
B) a+bab\dfrac{{a + b}}{{ab}}
C) a−bab\dfrac{{a - b}}{{ab}}
D) aba−b\dfrac{{ab}}{{a - b}}

Explanation

Solution

Use the relation between the arithmetic progression and the harmonic progression, to solve the question. Arithmetic progression is the reciprocal of the harmonic progression. Then solve the question using the formula of arithmetic progression. Remember the difference between the two numbers of arithmetic progression is always the same. Subtract the first term from the second term, second term from the third term and third term from the fourth term. Then the subtraction of first term from second term is equal to subtraction of third term from fourth term. Solve the equation and you will get the answer.

Complete step by step answer:
Given, a, H1, H2, b
We know that Arithmetic progression is the reciprocal of the harmonic progression.
Therefore, 1a,1H1,1H2,1b\dfrac{1}{a},\dfrac{1}{{{H_1}}},\dfrac{1}{{{H_2}}},\dfrac{1}{b}
We also know that,
The difference between the two consecutive numbers is always the same in arithmetic progression.
Therefore, 1H1−1a=\dfrac{1}{{{H_1}}} - \dfrac{1}{a} = 1H2−1H1=\dfrac{1}{{{H_2}}} - \dfrac{1}{{{H_1}}} = 1b−1H2\dfrac{1}{b} - \dfrac{1}{{{H_2}}}
Now, we can say that
1H1−1a=\dfrac{1}{{{H_1}}} - \dfrac{1}{a} = 1b−1H2\dfrac{1}{b} - \dfrac{1}{{{H_2}}}
Therefore,
1H1+1H2=1a+1b\dfrac{1}{{{H_1}}} + \dfrac{1}{{{H_2}}} = \dfrac{1}{a} + \dfrac{1}{b}
On solving by taking the L.C.M (least common factor) of two numbers
H1+H2H1.H2=a+bab\dfrac{{{H_1} + {H_2}}}{{{H_1}.{H_2}}} = \dfrac{{a + b}}{{ab}}

Therefore, the correct answer is a+bab\dfrac{{a + b}}{{ab}}.

Additional Information: Remember the difference between the two numbers of arithmetic progression is always the same. Subtract the first term from the second term, second term from the third term and third term from the fourth term. Then the subtraction of the first term from the second term is equal to the subtraction of the third term from the fourth term. Solve the equation and you will get the answer.

Note: You can also solve by simplifying the question first, Like this,
H1+H2H1.H2=\dfrac{{{H_1} + {H_2}}}{{{H_1}.{H_2}}} = H1H1.H2\dfrac{{{H_1}}}{{{H_1}.{H_2}}}+H2H1.H2\dfrac{{{H_2}}}{{{H_1}.{H_2}}}
H1+H2H1.H2=\dfrac{{{H_1} + {H_2}}}{{{H_1}.{H_2}}} = 1H2\dfrac{1}{{{H_2}}}+1H1\dfrac{1}{{{H_1}}}
Now, you can directly get the answer as 1H1+1H2=1a+1b\dfrac{1}{{{H_1}}} + \dfrac{1}{{{H_2}}} = \dfrac{1}{a} + \dfrac{1}{b}
On solving this, you will get
H1+H2H1.H2=a+bab\dfrac{{{H_1} + {H_2}}}{{{H_1}.{H_2}}} = \dfrac{{a + b}}{{ab}}