Question

If $b+c,c+a,a+b$ are in HP, then $a^2,b^2,c^2$ are in:

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

Answer 1

Answer: (A)

Since, $b+c,c+a,a+b$ are in HP
$\Rightarrow$ $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ are in AP
$\Rightarrow$ $\frac{1}{c+a}=\frac{1}{2}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)$
$\Rightarrow$ $\frac{2}{c+a}=\frac{a+b+b+c}{(a+b)(b+c)}$
$\Rightarrow$ $(a+2b+c)(a+c)=2(a+b)(b+c)$
$\Rightarrow$ $a^2+ac+2ab+2bc+ac+c^2$
$=2(ab+ac+b^2+bc)$
$\Rightarrow$ $a^2+c^2=2b^2$
$\Rightarrow$ $a^2,b^2,c^2$ are in AP