Question

If $\frac{(b+c-a)}{a},\frac{(c+a-b)}{b},\frac{(a+b-c)}{c}$ are in $AP$, then $a,b,c$ are in

• A. $G.P.$
• B. $A.P.$
• C. $H.P.$
• D. $A.G.P.$

Answer 1

The correct option is C

$H.P.$

Step 1. Finding the value of $a,b,c:$

Given, $\frac{(b+c-a)}{a},\frac{(c+a-b)}{b},\frac{(a+b-c)}{c}$ are in $AP$

Add $2$ to all terms.

$\Rightarrow \frac{\left(b+c-a\right)}{a}+2,\frac{\left(c+a-b\right)}{b}+2,\frac{\left(a+b-c\right)}{c}+2$ are in $AP.$

$\Rightarrow \frac{\left(b+c-a+2a\right)}{a},\frac{\left(c+a+b+2b\right)}{b},\frac{\left(a+b-c+2c\right)}{c}$ are in $AP.$

$\Rightarrow \frac{\left(a+b+c\right)}{a},\frac{\left(a+b+c\right)}{b},\frac{\left(a+b+c\right)}{c}$ are in $AP.$

Step 2. Divide all terms by $\left(a+b+c\right)$.

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

$\mathbf{\therefore }\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{}$ are in $\mathit{H}\mathit{P}\mathbf{.}$

Hence, option(C) is correct.