Question

When \[\frac{1}{a} + \frac{1}{c} + \frac{1}{{a - b}} + \frac{1}{{c - b}} = 0\] and \[b \ne a \ne d\], then \[a,b,c\] are
A. In H.P.
B. In G.P.
C. In A.P.
D. None of these

Answer 1

Hint:
By using the LCM of the denominators, we will add the given pattern in this question before solving it. We have been informed that \[b \ne a + c\], to answer this question, we'll use the formula for harmonic progression and its definition. If the sequence mentioned above is in H.P., the outcome will be as follows: \[\frac{2}{b} = \frac{1}{a} + \frac{1}{c}\]
Complete step-by-step solution
We are aware that a harmonic progression is a series of real numbers that is created by taking the reciprocals of arithmetic progressions without a zero.
Here, we have been given in the question that,
\[\frac{1}{a} + \frac{1}{c} + \frac{1}{{a - b}} + \frac{1}{{c - b}} = 0\]
Now, we have to rearrange the terms, we get
\[\frac{1}{a} + \frac{1}{{c - b}} + \frac{1}{c} + \frac{1}{{a - b}} = 0\]
Now, we have to add the first two terms and then the other two terms, we obtain
\[\frac{{a + c - b}}{{a(c - b)}} + \frac{{a + c - b}}{{c(a - b)}} = 0\]
We have been given that \[b \ne a + c\], so write as \[a + c - b \ne 0\]
Now, we have to divide the fractions in the above equation by \[a + c - b\]:
\[\frac{{a + c - b}}{{a(c - b)(a + c - b)}} + \frac{{a + c - b}}{{c(a - b)(a + c - b)}} = 0\]
On eliminating the similar terms, it gives:
\[\frac{1}{{a(c - b)}} + \frac{1}{{c(a - b)}} = 0\]
We can write it as follows by moving one term to the RHS:
\[\frac{1}{{a(c - b)}} = - \frac{1}{{c(a - b)}}\]
Now, we have to cross multiply the above equation, we get
\[a(c - b) = - \{ c(a - b)\} \]
On breaking the terms in the above equation, it gives:
\[ac - ba = - ac + bc\]
Now, we have to arrange the similar terms, we get
\[ac + ac = bc + ab\]
The above equation gives;
\[2ac = bc + ab\]
The above equation should be divided by \[abc\] on either sides of the equation:
\[\frac{{2ac}}{{abc}} = \frac{{bc}}{{abc}} + \frac{{ab}}{{abc}}\]
A new equation is obtained:
\[\frac{2}{b} = \frac{1}{a} + \frac{1}{c}\]
The above obtained equation satisfies the condition of H.P.
Hence, the option A is correct.
Note:
We should be aware that if letters a, b, and c are in the HP, then their opposites, or the arithmetic progression, should be in the AP. The relationship above can also be expressed as \[\frac{{a + c}}{{ac}} = \frac{2}{b}\]. Alternatively, cross multiplication yields \[b(a + c) = 2ac\].