Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in which series Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in
(1)AP
(2) GP
(3) HP
(4) none of these
My approach is as follow
$2{b^2} = {a^2} + {c^2} \Rightarrow 2{b^2} + {b^2} = {a^2} + {c^2} + {b^2} \Rightarrow 3{b^2} = {a^2} + {c^2} + {b^2}$
${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$
${\left( {a + b + c} \right)^2} = 3{b^2} + 2 \Rightarrow {\left( {a + b + c} \right)^2} - {b^2} = 2{b^2} + 2$
$\Rightarrow \left( {a + b + c - b} \right)\left( {a + b + c + b} \right) = 2{b^2} + 2$
$\Rightarrow \left( {a + c} \right)\left( {a + c + 2b} \right) = 2{b^2} + 2 \Rightarrow {\left( {a + c} \right)^2} + 2b\left( {a + c} \right) = 2{b^2} + 2$
$ \Rightarrow {\left( {a + c} \right)^2} + 2\left( {ab + bc} \right) = 2{b^2} + 2 \Rightarrow {\left( {a + c} \right)^2} + 2\left( {1 - ac} \right) = 2{b^2} + 2$
$\Rightarrow {\left( {a + c} \right)^2} - 2ac = 2{b^2} \Rightarrow {\left( {a + c} \right)^2} = 2\left( {{b^2} + ac} \right)$
$ \Rightarrow \frac{{a + c}}{{{b^2} + ac}} = \frac{2}{{\left( {a + c} \right)}}$
Not able to proceed further.

$$b^2-a^2=c^2-b^2=k(\ne0)\text{say}\implies b^2=a^2+k, c^2-a^2=2k$$
$$(c-b)(c+b)=k\iff b+c=\dfrac k{c-b}, \text{similarly } a+b=\dfrac k{b-a}, c-a=\dfrac{2k}{c+a}$$
$$\dfrac1{b+c}+\dfrac1{a+b}=\dfrac{c-b+b-a}k=\dfrac{2k}{k(c+a)}=? $$
