Question

If $1/(b-a)+1/(b-c)=(1/a)+(1/c),$then $a,b,c$are in

• A. $AP$
• B. $GP$
• C. $HP$
• D. none of these

Answer 1

The correct option is C

$HP$

Step 1, Finding the relation:

If $a,b,c$ are in $HP.$

Then $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in $AP.$

$\Rightarrow \left(\frac{1}{a}\right)+\left(\frac{1}{c}\right)=\frac{2}{b}...\left(i\right)$

Let $d$ be the common difference of $AP.$

$\begin{array}{rcl}\left(\frac{1}{b}\right)-\left(\frac{1}{a}\right)& =& \left(\frac{1}{c}\right)-\left(\frac{1}{b}\right)\\ & =& d\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{\left(a-b\right)}{ab}& =& \frac{b-c}{bc}\\ & =& d\end{array}$

$\Rightarrow$ $a-b=abd...\left(ii\right)$ and

$$\begin{gathered} b-c=bcd...\left(iii\right) \\ LHS=\frac{-1}{\left(a-b\right)}+\frac{1}{\left(b-c\right)} \\ =\left(\frac{-1}{abd}\right)+\left(\frac{1}{bcd}\right) \end{gathered}$$

from (ii) and (iii), we get

$$\begin{gathered} =\left(\frac{1}{bd}\right)\left(\frac{1}{c}-\frac{1}{a}\right) \\ =\left(\frac{1}{bd}\right)2d\left[\because \frac{1}{c}-\frac{1}{a}=2d\right] \\ =\frac{2}{b} \\ =\left(\frac{1}{a}\right)+\left(\frac{1}{c}\right)\left(from(i\right)) \end{gathered}$$

$=$RHS

Hence verified

Hence, option(C) is correct.