Question

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

Answer 1

Hint: 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 $a$, $b$, $c$ in A.P and calculate the value of $b$. After getting the value of $b$ we will substitute the value of $b$ 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,
$\dfrac{1}{a}+\dfrac{1}{a-b}+\dfrac{1}{c}+\dfrac{1}{c-b}=0$
Simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow -\dfrac{1}{a-b}-\dfrac{1}{c-b}=\dfrac{1}{a}+\dfrac{1}{c} \\
 & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{a}+\dfrac{1}{c} \\
\end{align}$
Consider the L.H.S part L.H.S $=\dfrac{1}{b-a}+\dfrac{1}{b-c}$.
If $a$, $b$, $c$ are in A.P, then common difference is given by $d=b-a=c-b$.
Now substituting the value $b-a=c-b$ in L.H.S, then we will get
$\begin{align}
  & \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{c-b}+\dfrac{1}{b-c} \\
 & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{c-b}-\dfrac{1}{c-b} \\
 & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=0 \\
\end{align}$
$\therefore $L.H.S $\ne $ R.H.S i.e. $a$, $b$, $c$ are not in A.P.
If $a$, $b$, $c$ are in H.P, then we must have $\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$.
$\begin{align}
  & \Rightarrow \dfrac{2}{b}=\dfrac{c+a}{ac} \\
 & \Rightarrow 2ac=(c+a)b \\
 & \Rightarrow b=\dfrac{2ac}{c+a} \\
\end{align}$
Substituting the $b$ value in L.H.S, then we will get
$\begin{align}
  & \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{\dfrac{2ac}{c+a}-a}+\dfrac{1}{\dfrac{2ac}{c+a}-c} \\
 & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{1}{\dfrac{2ac-(c+a)a}{c+a}}+\dfrac{1}{\dfrac{2ac-(c+a)c}{c+a}} \\
 & \Rightarrow \dfrac{1}{b-a}+\dfrac{1}{b-c}=\dfrac{c+a}{a(2c-(c+a))}+\dfrac{c+a}{c(2a-(c+a))} \\
\end{align}$

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