STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Then ${b^2} - {a^2} = {c^2} - {b^2}$

$ \Rightarrow $ $(b - a)(b + a) = (c - b)(c + b)$

==> $\frac{{b - a}}{{c + b}} = \frac{{c - b}}{{b + a}}$

==> $\frac{{(b - a)(a + b + c)}}{{(c + a)(b + c)}} = \frac{{(c - b)(a + b + c)}}{{(a + b)(c + a)}}$

$ \Rightarrow $ $\frac{{{b^2} + bc - ac - {a^2}}}{{(c + a)(b + c)}} = \frac{{{c^2} + ac - ab - {b^2}}}{{(a + b)(c + a)}}$

$ \Rightarrow $ $\frac{b}{{c + a}} - \frac{a}{{b + c}} = \frac{c}{{a + b}} - \frac{b}{{c + a}}$

Hence $\frac{a}{{b + c}},\;\frac{b}{{c + a}},\;\frac{c}{{a + b}}$ be in $A.P.$