Question

When $\dfrac{1}{a} + \dfrac{1}{c} + \dfrac{1}{{a - b}} + \dfrac{1}{{c - b}} = 0$ and $b \ne a + c,$ then $a,b,c$ are in
A) AP
B) GP
C) HP
D) None of these

Answer 1

In this question we will add the given pattern by taking the LCM of the denominators and then we solve them . Here we have been given that
$b \ne a + c,$ We will use the definition of Harmonic progression and its formula to solve this question.
If the above given sequence is in H.P then, we will get the result in the following form:
$\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$

Complete step by step solution:
We know that harmonic progression is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic progressions that does not contain zero.
Here we have been given that
$\dfrac{1}{a} + \dfrac{1}{c} + \dfrac{1}{{a - b}} + \dfrac{1}{{c - b}} = 0$
We will arrange the terms as
$\dfrac{1}{a} + \dfrac{1}{{c - b}} + \dfrac{1}{c} + \dfrac{1}{{a - b}} = 0$
Now we will add the first two terms and the other two terms, so we have
$\dfrac{{a + c - b}}{{a(c - b)}} + \dfrac{{a + c - b}}{{c(a - b)}} = 0$
We have been given that
$b \ne a + c,$so from this we can also write this as
$a + c - b \ne 0$
Now we will divide the dfractions by $a + c - b$,
$\dfrac{{a + c - b}}{{a(c - b)(a + c - b)}} + \dfrac{{a + c - b}}{{c(a - b)(a + c - b)}} = 0$
We will eliminate the similar terms, so it gives:
$\dfrac{1}{{a(c - b)}} + \dfrac{1}{{c(a - b)}} = 0$
By taking one term to the RHS, we can write it as :
$\dfrac{1}{{a(c - b)}} = - \dfrac{1}{{c(a - b)}}$
On cross multiplication, it gives:
a(c - b) = - \left\\{ {c(a - b)} \right\\}
By breaking the terms, we have;
$ac - bc = - ac + ab$
We will arrange the similar terms together:
$ac + ac = bc + ab$
It gives
$2ac = bc + ab$
We can divide both the sides by $abc$, so it gives:
$\dfrac{{2ac}}{{abc}} = \dfrac{{bc}}{{abc}} + \dfrac{{ab}}{{abc}}$
It gives us the new equation i.e.
$\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$
**Now we can see that this is the important condition of the Harmonic progression.
Hence the correct option is option (C) HP **

Note:
We should note that if $a,b,c$ are in HP, then the reciprocal of them should be in AP i.e. Arithmetic progression. We can also write the above relation as
$\dfrac{{a + c}}{{ac}} = \dfrac{2}{b}$
Or by cross multiplication it gives,
$b(a + c) = 2ac$