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

Hint:Here we basically rearrange the equation
\[\dfrac{1}{a} + \dfrac{1}{c} + \dfrac{1}{{(a - b)}} + \dfrac{1}{{(c - b)}} = 0\]
In such a way we can substitute \[b \ne a + c\] in it and further simplify the equation to obtain the required answer.
The sequence of real numbers that are reciprocals of an arithmetic progression that does not contain 0 is known as Harmonic progression.

Formula used:
If \[a,b,c\] are in harmonic progression, then \[\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}\].

Complete step by step Solution:
Here the given equations are
\[\dfrac{1}{a} + \dfrac{1}{c} + \dfrac{1}{{(a - b)}} + \dfrac{1}{{(c - b)}} = 0\]
\[b \ne a + c\]
Next, we rearrange the terms of the equation \[\dfrac{1}{a} + \dfrac{1}{c} + \dfrac{1}{{(a - b)}} + \dfrac{1}{{(c - b)}} = 0\] in the following way,
\[\left[\left(\dfrac{1}{a}\right) + \dfrac{1}{{(c - b)}}\right] +\left [\left(\dfrac{1}{c}\right) + \dfrac{1}{{(a - b)}}\right] = 0\]
Adding the fractional values with the brackets we get
\[\left[\dfrac{{(c - b + a)}}{{a(c - b)}}\right] +\left [\dfrac{{(a - b + c)}}{{c(a - b)}}\right] = 0\]
Next, we take \[(a + c - b)\] as common,
\[(a + c - b)\left(\dfrac{1}{{a(c - b)}} + \dfrac{1}{{c(a - b)}}\right) = 0\]
Further, as we are given \[b \ne a + c\], we rearrange the terms to get
\[(a + c - b) \ne 0\]
Since one factor of given equation is \[(a + c - b) \ne 0\], so the another factor of the equation must be equal to zero.
So \[\dfrac{1}{{a(c - b)}} + \dfrac{1}{{c(a - b)}} = 0\]
Solving fractions within the brackets, we get
\[\dfrac{{[c(a - b) + a(c - b)]}}{{ac(a - b)(c-b)}} = 0\]
Sending denominator to the other side of the equals to sign we get
\[[c(a - b) + a(c - b)] = 0\]
Simplifying further we get
\[ca - cb + ac - ab = 0\]
\[2ac = bc + ab\]
Now we divide the equation by \[abc\] on both sides
\[\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}\]
Which is necessary condition to prove that a, b and c are in HP
Hence proved \[a,b,c\] are HP.

Hence, option C is the correct answer.

Note: Student often confused with geometric progression and harmonic progression. But harmonic progression is reciprocal of terms in arithmetic progression whereas geometric progression has different relation between terms.