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.

$a^2, b^2, c^2$ are in AP.
Then $a^2+1,b^2+1,c^2+1$ are also in AP.
$a^2+ab+bc+ca, b^2+ab+bc+ca, c^2+ab+bc+ca$ are in AP.
$\begin{align}a^2+ab+bc+ca&=a(a+b) +c(b+a)\\&=(a+c) (a+b) \end{align}$
$\begin{align}b^2+ab+bc+ca&=b(b+a) +c(b+a)\\&=(b+c) (a+b) \end{align}$
$\begin{align}c^2+ab+bc+ca&=c(c+a) +b(c+a)\\&=(b+c) (c+a) \end{align}$
Now divide by $(a+b) (b+c) (c+a) $.
Hence $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in AP.

 $(b+c), (c+a), (a+b) $ are in HP.

