# If $A$ be the A.M. and $H$ be the H.M. between two numbers $a$ and $b$, then show that

$\dfrac{{a - A}}{{a - H}} \times \dfrac{{b - A}}{{b - H}} = \dfrac{A}{H}$

Answer

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Hint- Simply put the general values of A.M. and H.M. in the L.H.S. of the equation and manipulate some terms.

We know that

$\because A$ is the A.M. of the numbers $a, b$

\[\therefore A = \dfrac{{a + b}}{2}\]

$\because H$ is the H.M. of the numbers $a, b$

$\therefore H = \dfrac{{2ab}}{{a + b}}$

Now substituting this in the L.H.S. of the given question

$

\Rightarrow \dfrac{{a - A}}{{a - H}} \times \dfrac{{b - A}}{{b - H}} \\

\Rightarrow \dfrac{{a - \dfrac{{a + b}}{2}}}{{a - \dfrac{{2ab}}{{a + b}}}} \times \dfrac{{b - \dfrac{{a + b}}{2}}}{{b - \dfrac{{2ab}}{{a + b}}}} \\

$

Further solving the numerator and denominator by taking LCM

$

\Rightarrow \dfrac{{\dfrac{{2a - a - b}}{2}}}{{\dfrac{{{a^2} + ab - 2ab}}{{a + b}}}} \times \dfrac{{\dfrac{{2b - a - b}}{2}}}{{\dfrac{{{b^2} + ab - 2ab}}{{a + b}}}} \\

\Rightarrow \dfrac{{\dfrac{{a - b}}{2}}}{{\dfrac{{{a^2} - ab}}{{a + b}}}} \times \dfrac{{\dfrac{{b - a}}{2}}}{{\dfrac{{{b^2} - ab}}{{a + b}}}} \\

\Rightarrow \dfrac{{\left( {a - b} \right)\left( {a + b} \right)}}{{2\left( {{a^2} - ab} \right)}} \times \dfrac{{\left( {b - a} \right)\left( {a + b} \right)}}{{2\left( {{b^2} - ab} \right)}} \\

\Rightarrow \dfrac{{\left( {a - b} \right)\left( {a + b} \right)}}{{2a\left( {a - b} \right)}} \times \dfrac{{\left( {b - a} \right)\left( {a + b} \right)}}{{2b\left( {b - a} \right)}} \\

$

Now cancelling the common terms from the numerator and denominator

$ \Rightarrow \dfrac{{a + b}}{{2a}} \times \dfrac{{a + b}}{{2b}}$

Above equation can be manipulated as done below in order to obtain the R.H.S.

$

\Rightarrow \dfrac{{a + b}}{2} \times \dfrac{{a + b}}{{2ab}} \\

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

$

As we know that

$

\because \dfrac{{a + b}}{2} = A{\text{ and }}\dfrac{{2ab}}{{a + b}} = H \\

\Rightarrow \dfrac{A}{H} \\

$

It becomes equal to the R.H.S.

Hence, the equation in the question is proved.

Note- Arithmetic mean and harmonic mean are one of the most important concepts and are used in almost all types of problems. Arithmetic mean represents a number that is obtained by dividing the sum of the elements of a set by the number of values in the set. The harmonic mean is defined as the reciprocal of the arithmetic mean of the given data values. It is based on all values present in the set.

We know that

$\because A$ is the A.M. of the numbers $a, b$

\[\therefore A = \dfrac{{a + b}}{2}\]

$\because H$ is the H.M. of the numbers $a, b$

$\therefore H = \dfrac{{2ab}}{{a + b}}$

Now substituting this in the L.H.S. of the given question

$

\Rightarrow \dfrac{{a - A}}{{a - H}} \times \dfrac{{b - A}}{{b - H}} \\

\Rightarrow \dfrac{{a - \dfrac{{a + b}}{2}}}{{a - \dfrac{{2ab}}{{a + b}}}} \times \dfrac{{b - \dfrac{{a + b}}{2}}}{{b - \dfrac{{2ab}}{{a + b}}}} \\

$

Further solving the numerator and denominator by taking LCM

$

\Rightarrow \dfrac{{\dfrac{{2a - a - b}}{2}}}{{\dfrac{{{a^2} + ab - 2ab}}{{a + b}}}} \times \dfrac{{\dfrac{{2b - a - b}}{2}}}{{\dfrac{{{b^2} + ab - 2ab}}{{a + b}}}} \\

\Rightarrow \dfrac{{\dfrac{{a - b}}{2}}}{{\dfrac{{{a^2} - ab}}{{a + b}}}} \times \dfrac{{\dfrac{{b - a}}{2}}}{{\dfrac{{{b^2} - ab}}{{a + b}}}} \\

\Rightarrow \dfrac{{\left( {a - b} \right)\left( {a + b} \right)}}{{2\left( {{a^2} - ab} \right)}} \times \dfrac{{\left( {b - a} \right)\left( {a + b} \right)}}{{2\left( {{b^2} - ab} \right)}} \\

\Rightarrow \dfrac{{\left( {a - b} \right)\left( {a + b} \right)}}{{2a\left( {a - b} \right)}} \times \dfrac{{\left( {b - a} \right)\left( {a + b} \right)}}{{2b\left( {b - a} \right)}} \\

$

Now cancelling the common terms from the numerator and denominator

$ \Rightarrow \dfrac{{a + b}}{{2a}} \times \dfrac{{a + b}}{{2b}}$

Above equation can be manipulated as done below in order to obtain the R.H.S.

$

\Rightarrow \dfrac{{a + b}}{2} \times \dfrac{{a + b}}{{2ab}} \\

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

$

As we know that

$

\because \dfrac{{a + b}}{2} = A{\text{ and }}\dfrac{{2ab}}{{a + b}} = H \\

\Rightarrow \dfrac{A}{H} \\

$

It becomes equal to the R.H.S.

Hence, the equation in the question is proved.

Note- Arithmetic mean and harmonic mean are one of the most important concepts and are used in almost all types of problems. Arithmetic mean represents a number that is obtained by dividing the sum of the elements of a set by the number of values in the set. The harmonic mean is defined as the reciprocal of the arithmetic mean of the given data values. It is based on all values present in the set.

Last updated date: 20th Sep 2023

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