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Question: If we have an expression as \(\dfrac{1}{a}+\dfrac{1}{a-b}+\dfrac{1}{c}+\dfrac{1}{c-b}=0\) and \(a+c\...

If we have an expression as 1a+1a−b+1c+1c−b=0\dfrac{1}{a}+\dfrac{1}{a-b}+\dfrac{1}{c}+\dfrac{1}{c-b}=0 and a+c≠ba+c\ne b then aa, bb, cc are in
(A). A.P
(B). H.P
(C). G.P
(D). None of these

Explanation

Solution

We will first simplify the given equation so that the terms having relation between the variables in L.H.S and the remaining are in R.H.S. Now we will consider the L.H.S part and take the condition if aa, bb, cc in A.P and calculate the value of bb. After getting the value of bb we will substitute the value of bb in the L.H.S and check whether the result is equal to R.H.S or not. Further, we will do the same process for remaining options like H.P and G.P and check whether the value is equal to R.H.S are not.

Complete step-by-step solution:
Given that,
1a+1a−b+1c+1c−b=0\dfrac{1}{a}+\dfrac{1}{a-b}+\dfrac{1}{c}+\dfrac{1}{c-b}=0
Simplifying the above equation, then we will get
⇒−1a−b−1c−b=1a+1c ⇒1b−a+1b−c=1a+1c \begin{aligned} & \Rightarrow -\dfrac{1}{a-b}-\dfrac{1}{c-b}=\dfrac{1}{a}+\dfrac{1}{c} \\\ & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c} \\\ \end{aligned}
Consider the L.H.S part L.H.S =1b−a+1b−c=\dfrac{1}{b-a}+\dfrac{1}{b-c}.
If aa, bb, cc are in A.P, then common difference is given by d=b−a=c−bd=b-a=c-b.
Now substituting the value b−a=c−bb-a=c-b in L.H.S, then we will get
1b−a+1b−c=1c−b+1b−c ⇒1b−a+1b−c=1c−b−1c−b ⇒1b−a+1b−c=0 \begin{aligned} & \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{c-b}+\dfrac{1}{b-c} \\\ & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{c-b}-\dfrac{1}{c-b} \\\ & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=0 \\\ \end{aligned}
∴\therefore L.H.S ≠\ne R.H.S i.e. aa, bb, cc are not in A.P.
If aa, bb, cc are in H.P, then we must have 2b=1a+1c\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}.
⇒2b=c+aac ⇒2ac=(c+a)b ⇒b=2acc+a \begin{aligned} & \Rightarrow \dfrac{2}{b}=\dfrac{c+a}{ac} \\\ & \Rightarrow 2ac=(c+a)b \\\ & \Rightarrow b=\dfrac{2ac}{c+a} \\\ \end{aligned}
Substituting the bb value in L.H.S, then we will get
1b−a+1b−c=12acc+a−a+12acc+a−c ⇒1b−a+1b−c=12ac−(c+a)ac+a+12ac−(c+a)cc+a ⇒1b−a+1b−c=c+aa(2c−(c+a))+c+ac(2a−(c+a)) \begin{aligned} & \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{\dfrac{2ac}{c+a}-a}+\dfrac{1}{\dfrac{2ac}{c+a}-c} \\\ & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{\dfrac{2ac-(c+a)a}{c+a}}+\dfrac{1}{\dfrac{2ac-(c+a)c}{c+a}} \\\ & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{c+a}{a(2c-(c+a))}+\dfrac{c+a}{c(2a-(c+a))} \\\ \end{aligned}

Taking c+ac+a common in the above equation, then we will get
⇒1b−a+1b−c=(c+a)[1a(c−a)+1c(a−c)]\Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=(c+a)\left[ \dfrac{1}{a(c-a)}+\dfrac{1}{c(a-c)} \right]
Now multiply minus to the term (a−c)\left( a-c \right) in the above equation, then we will get
⇒1b−a+1b−c=(c+a)[1a(c−a)−1c(c−a)]\Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=(c+a)\left[ \dfrac{1}{a(c-a)}-\dfrac{1}{c(c-a)} \right]
Taking (c−a)\left( c-a \right) common from the above equation, then we will get
⇒1b−a+1b−c=(c+a)(c−a)[1a−1c]\Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{(c+a)}{(c-a)}\left[ \dfrac{1}{a}-\dfrac{1}{c} \right]
Simplifying the above equation, then we will have
⇒1b−a+1b−c=(c+a)(c−a)[c−aac] ⇒1b−a+1b−c=c+aac ⇒1b−a+1b−c=cac+aac ⇒1b−a+1b−c=1a+1c \begin{aligned} & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{(c+a)}{(c-a)}\left[ \dfrac{c-a}{ac} \right] \\\ & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{c+a}{ac} \\\ & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{c}{ac}+\dfrac{a}{ac} \\\ & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c} \\\ \end{aligned}
Here we got L.H.S == R.H.S. So that we can write aa, bb, cc are H.P.

Note: For this problem we got the answer at the second option. So, we will terminate the problem. If you do not get the answer in the second option, then you must go for the third option by using the condition b=acb=\sqrt{ac}.