Show that a², b² and c² are in A.P., if

$\dfrac{1}{b + c}, \dfrac{1}{c + a} \text{ and } \dfrac{1}{a + b}$ are in A.P.

Since $\dfrac{1}{b + c}, \dfrac{1}{c + a}, \dfrac{1}{a + b}$ are in A.P.,

$\Rightarrow \dfrac{1}{c + a} - \dfrac{1}{b + c} = \dfrac{1}{a + b} - \dfrac{1}{c + a} \\[1em] \Rightarrow \dfrac{(b + c) - (c + a)}{(c + a)(b + c)} = \dfrac{(c + a) - (a + b)}{(a + b)(c + a)} \\[1em] \Rightarrow \dfrac{b - a}{(c + a)(b + c)} = \dfrac{c - b}{(a + b)(c + a)} \\[1em] \Rightarrow \dfrac{b - a}{b + c} = \dfrac{c - b}{a + b} \\[1em]$

⇒ (b − a)(a + b) = (c − b)(b + c)

⇒ ab + b² − a² − ab = bc + c² − b² − bc

⇒ b² − a² = c² − b²

⇒ b² − a² = c² − b²

We have,

b² − a² = c² − b²

Since consecutive differences are equal,

Therefore,

a², b², c² are in A.P.

Hence, a², b², c² are in A.P.