Question

Show that a², b², c² are in A.P., if \[\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}\] are in A.P.

Answer 1

\[\frac{1}{b + c}, \frac{1}{c + a}\] and \[\frac{1}{a + b}\] are in A.P.

⇒ \[\frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b} - \frac{1}{c + a}\]

⇒ \[\frac{b + c - c - a}{(c + a)(b + c)} = \frac{c + a - a - b}{(a + b)(c + a)}\]

⇒ \[\frac{b - a}{b + c} = \frac{c - b}{a + b}\]

⇒ \[b^2 - a^2 = c^2 - b^2\]

⇒ \[2b^2 = a^2 + c^2\]

⇒ \[a^2\], \[b^2\] and \[c^2\] are in A.P.