Question

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

• A. $a,b,c$ are in AP
• B. $c,a,b$ are in AP
• C. $a^2,b^2,c^2$ are in AP
• D. $a,b,c$ are in GP

Answer 1

The correct option is C

$a^2,b^2,c^2$ are in AP

Explanation for the correct option:

Step 1. $\frac{a}{(b+c)},\frac{b}{(c+a)},\frac{c}{(a+b)}$ are in AP

$\Rightarrow$ $\frac{b}{(c+a)}-\frac{a}{(b+c)}=\frac{c}{(a+b)}-\frac{b}{(c+a)}$

$\Rightarrow$$\frac{(b^2+bc-ac-a^2)}{(c+a)(b+c)}=\frac{(c^2+ac-ab-b^2)}{(a+b)(c+a)}$

Step 2. Eliminate $(c+a)$ on both side and rearrange the numerator:

$\frac{(b^2-a^2+bc-ac)}{(b+c)}=\frac{(c^2-b^2+ac-ab)}{(a+b)}$

$\Rightarrow$$\frac{(b-a)(b+a)+c(b-a)}{(b+c)}=\frac{(c-b)(c+b)+a(c-b)}{(a+b)}$

Step 3. Take $(b-a)$ common on LHS and $(c-b)$ on RHS:

$\frac{(b-a)(b+a+c)}{(b+c)}=\frac{(c-b)(c+b+a)}{(a+b)}$

$\Rightarrow$ $\frac{b-a}{b+c}=\frac{c-b}{a+b}$

Step 4. By Cross multiplying, we get

$(ba-a^2+b^2-ab)=bc+c^2-b^2+bc$

$\Rightarrow$ $b^2-a^2=c^2-b^2$

$\Rightarrow$ $2b^2=c^2+a^2$

Thus, $a^2,b^2,c^2$ are in A.P

Hence, Option ‘C’ is Correct.