If {a^2},,{b^2},,{c^2} be in A.P., then frac{ | vedclass

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Question

(a) Since ${a^2},\;{b^2},\;{c^2}$ be in $A.P.$

Then ${b^2} - {a^2} = {c^2} - {b^2}$

$ \Rightarrow $ $(b - a)(b + a) = (c - b)(c + b)$

==>

Vedclass · Accepted Answer

$A.P.$

Vedclass · Answer

(a) Since ${a^2},\;{b^2},\;{c^2}$ be in $A.P.$

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 $A.P.$

If ${a^2},\,{b^2},\,{c^2}$ be in $A.P.$, then $\frac{a}{{b + c}},\,\frac{b}{{c + a}},\,\frac{c}{{a + b}}$ will be in

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