Question

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

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

Answer 1

This problem deals with the arithmetic mean and the harmonic mean. So in order to solve this problem we should have some knowledge about how to calculate the arithmetic mean of the given two numbers and also the harmonic mean of the two numbers.

Here the arithmetic mean of the two numbers a and b is given by:

$\Rightarrow A = \dfrac{{a + b}}{2}$

The harmonic mean of the two numbers a and b is given by:

$\Rightarrow \dfrac{1}{H} = \dfrac{1}{2}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)$

$\Rightarrow \dfrac{1}{H} = \dfrac{1}{2}\left( {\dfrac{{a + b}}{{ab}}} \right) = \dfrac{{a + b}}{{2ab}}$

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

Complete step-by-step solution:

Given that A is the A.M. Here A.M is the arithmetic mean

Also given that H is the H.M. Where H.M. is the harmonic mean

The arithmetic mean of a and b is given by:

$\Rightarrow A = \dfrac{{a + b}}{2}$

The harmonic mean of a and b is given by:

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

Now consider the left hand side of the given expression, $\dfrac{{a - A}}{{a - H}} \times \dfrac{{b - A}}{{b - H}} = \dfrac{A}{H}$, as given below;

Here L.H.S is $\dfrac{{a - A}}{{a - H}} \times \dfrac{{b - A}}{{b - H}},$ now substitute the values of $A$ and $H$, in this expression as given below:

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

Now simplifying the numerators and the denominators of the above expression by L.C.M. as given below

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

Grouping the like terms and unlike terms together in the numerators and denominators of the above expression, as given below:

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

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

The term $(a - b)$ and $(b - a)$ are cancelled in the numerators and denominators of the above expression, as given below:

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

$\Rightarrow \dfrac{{{{(a + b)}^2}}}{{4ab}}$, which is the L.H.S of the expression.

Now consider the R.H.S of the expression which is $\dfrac{A}{H}$, and substitute the values of $A$ and $H$, in this expression as given below:

$\Rightarrow \dfrac{A}{H} = \dfrac{{\left( {\dfrac{{a + b}}{2}} \right)}}{{\left( {\dfrac{{2ab}}{{a + b}}} \right)}}$

$\Rightarrow \dfrac{A}{H} = \dfrac{{{{(a + b)}^2}}}{{2(2ab)}}$

$\Rightarrow \dfrac{A}{H} = \dfrac{{{{(a + b)}^2}}}{{4ab}}$, which is the R.H.S of the expression.

$\therefore \dfrac{{a - A}}{{a - H}} \times \dfrac{{b - A}}{{b - H}} = \dfrac{{{{(a + b)}^2}}}{{4ab}}$ and

$\therefore \dfrac{A}{H} = \dfrac{{{{(a + b)}^2}}}{{4ab}}$

Thus L.H.S = R.H.S

Hence proved.

L.H.S = R.H.S = $\dfrac{{{{(a + b)}^2}}}{{4ab}}$

Note: Please note that this problem deals with the topic of progressions, where there are three types of progressions, which are arithmetic progression, geometric progression and harmonic progression. Here throughout the A.P there is a common difference $d$ and the arithmetic mean of any two numbers is given by $A = \dfrac{{a + b}}{2}$. Now the G.P has a common ratio of $r$, the geometric mean of any two numbers is given by $G = \sqrt {ab}$. We already discussed that the harmonic mean is given by $H = \dfrac{{2ab}}{{a + b}}$.