If $a^2,b^2,c^2\;$ are in AP ,$\frac{1}{b+c},\frac{1}{a+c},\frac{1}{b+a}\;$ are in

$(a)\;AP\qquad(b)\;HP\qquad(C)\;GP\qquad(d)\;None\;of\;the\;above$

Explanation : $check\;if\;\frac{1}{b+c},\frac{1}{a+c},\frac{1}{b+a} \;are\;in\;AP$
$\frac{1}{a+c}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}$
$\frac{b-a}{(a+c)(b+c)}=\frac{c-b}{(a+b)(a+c)}$
$\frac{b-a}{b+c}=\frac{c-b}{a+b}$
$b^2-a^2=c^2-b^2\quad\;This\;is\;true\;as\;a^2,b^2\xi\;c^2 \;are\;in\;AP$
Hence order assumption that $\frac{1}{b+c},\frac{1}{a+c},\frac{1}{a+b}\;$are in AP is true.