Question

When $\frac{1}{a}+\frac{1}{c}+\frac{1}{\left(a-b\right)}+\frac{1}{\left(c-b\right)}=0$ and $b\ne a+c$, 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: simplify the given equation

The given equation is,

$\frac{1}{a}+\frac{1}{c}+\frac{1}{\left(a-b\right)}+\frac{1}{\left(c-b\right)}=0$

$\Rightarrow$ $\left[\frac{1}{a}+\frac{1}{\left(c-b\right)}\right]+\left[\frac{1}{c}+\frac{1}{\left(a-b\right)}\right]=0$

$\Rightarrow$ $\left[\frac{c-b+a}{a\left(c-b\right)}\right]+\left[\frac{a-b+c}{c\left(a-b\right)}\right]=0$

$\Rightarrow$ $\left(a+c-b\right)\left[\frac{1}{a\left(c-b\right)}+\frac{1}{c\left(a-b\right)}\right]=0$

Step 2: Apply the given condition $b\ne a+c$

Since, $b\ne a+c$. So $a+c-b\ne 0$

$\Rightarrow$ $\frac{1}{a\left(c-b\right)}+\frac{1}{c\left(a-b\right)}=0$

$\Rightarrow$ $\frac{c\left(a-b\right)+a\left(c-b\right)}{ac\left(c-b\right)\left(a-b\right)}=0$

Since the denominator cannot be $0$.

$\Rightarrow$ $c\left(a-b\right)+a\left(c-b\right)=0$

$\Rightarrow$ $ca-cb+ac-ab=0$

$\Rightarrow$ $2ac-bc-ab=0$

$\Rightarrow$ $2ac=bc+ab$

divide both sides by $abc$, we get

$\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

$\Rightarrow$ $a,b,c$ are in HP.

Hence, the correct option is $C)HP$.