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If a, b, c be in H.P, show that a: a-b= a + c: a-c

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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

$\dfrac{1}{b} = \dfrac{1}{a} + \dfrac{1}{c}$
Now it is been given that a, b, c are in H.P so,
Let’s simplify it to solve for b we get
${\text{b = }}\dfrac{{2ac}}{{a + c}}$…………………………………………… (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.

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