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If $a, b, c$ are in H.P, then show that $\dfrac{a}{a-b}=\dfrac{a+c}{a-c}$.

Last updated date: 15th Jul 2024
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Hint: Use the basic definition of three numbers being in H.P (harmonic progression)  which is if three numbers a, b and c are in H.P then we can say that
$\dfrac{1}{b} = \dfrac{1}{a} + \dfrac{1}{c}$ 
Which implies $b$ is
${\text{b = }}\dfrac{{2ac}}{{a + c}}$

Complete step by step solution:
Now it has been given that $a, b, c$ are in H.P 
${\text{b = }}\dfrac{{2ac}}{{a + c}}$…………………………………………… (1)
Now we have to prove that $\dfrac{a}{{a - b}} = \dfrac{{a + c}}{{a - c}}$
Considering the LHS,
We have $\dfrac{a}{{a - b}}$
Substitute value of $b$ from equation (1) then we have
$\dfrac{a}{{a - \dfrac{{2ac}}{{a + c}}}}$
Taking LCM in the denominator part we have 
$\dfrac{a}{{\dfrac{{{a^2} + ac - 2ac}}{{a + c}}}} = \dfrac{{a(a + c)}}{{{a^2} - ac}} = \dfrac{{a(a + c)}}{{a(a - c)}}$
This can be written as
$\dfrac{{a + c}}{{a - c}}$
Thus clearly LHS=RHS 
Hence proved.

Note: Whenever we come across such problems the key concept is to know the basic definitions of H.P, GP and AP. It will eventually help you get on the right track to reach the solution as all these have different definitions of three numbers to be in AP, GP or HP.