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8. Sequences and Series
medium
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
A
$A.P.$
B
$G.P.$
C
$H.P.$
D
None of these
Solution
(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 $A.P.$
Standard 11
Mathematics
