If (1)/(b+c), (1)/(c+a), (1)/(a+b) are in HP, then Which of the...

$\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in H.P. $\therefore b+c, c+a, a+b$ are in A.P. Subtract $a+b+c\Rightarrow-a,-b,-c$, are also in A.P. $\therefore a, b, c$, are also in A.P. So, statements 1 is correct. Let $a=a, b=a+d, c=a+2 d$$\begin{aligned} & \therefore\left[(c+a)^2\right]^2=(2 a+2 d)^4=16(a+d)^4 \\ & (b+c)^2 \cdot(a+b)^2=(2 a+3 d)^2(2 a+d)^2 \\ & \therefore(c+a)^4 \neq(b+c)^2 \cdot(a+b)^2 \end{aligned}$$$so,$(b+c)^2,(c+a)^2,(a+b)^2$ are not in G.P so, statements-2 is not correct.