Question

c^2/(a+b)^2, a^2/(b+c)^2,b^2/(a+c)^2 are in AP then show that (s-c) /(2s-c)^2, (s-a) /(2s-a) ^2,(s-b) /(2s-b) ^2 are also in AP

Answer 1

$\textbf{Given:}$

$\dfrac{c^2}{(a+b)^2},\dfrac{a^2}{(b+c)^2},\dfrac{b^2}{(c+a)^2}\;\text{are in A.P}$

$\textbf{To show:}$

$\dfrac{s-c}{(2s-c)^2},\dfrac{s-a}{(2s-a)^2},\dfrac{s-b}{(2s-b)^2}\;\text{are in A.P}$

$\textbf{Solution:}$

$\text{We use following two properties to prove the given problem}$

$\textbf{1.When a constant is subtracted from each term of an A.P, we a new A.P}$

$\textbf{2.When each term of an A.P is a divided by a non-zero constant, we a new A.P}$

$\text{Consider,}$

$\dfrac{c^2}{(a+b)^2},\dfrac{a^2}{(b+c)^2},\dfrac{b^2}{(c+a)^2}\;\text{are in A.P}$

$\text{Subtracting 1 from each term of the A.P, by properties of A.P}$

$\dfrac{c^2}{(a+b)^2}-1,\dfrac{a^2}{(b+c)^2}-1,\dfrac{b^2}{(c+a)^2}-1\;\text{are in A.P}$

$\dfrac{c^2-(a+b)^2}{(a+b)^2},\dfrac{a^2-(b+c)^2}{(b+c)^2},\dfrac{b^2-(c+a)^2}{(c+a)^2}\;\text{are in A.P}$

$\dfrac{(c-a-b)(c+a+b)}{(a+b)^2},\dfrac{(a-b-c)(a+b+c)}{(b+c)^2},\dfrac{(b-c-a)(b+c+a)}{(c+a)^2}\;\text{are in A.P}$

$\dfrac{(2c-a-b-c)(c+a+b)}{(a+b)^2},\dfrac{(2a-a-b-c)(a+b+c)}{(b+c)^2},\dfrac{(2b-c-a-b)(b+c+a)}{(c+a)^2}\;\text{are in A.P}$

$\dfrac{(2c-(a+b+c))(a+b+c)}{(a+b+c-c)^2},\dfrac{(2a-(a+b+c))(a+b+c)}{(a+b+c-a)^2},\\\dfrac{(2b-(a+b+c))(a+b+c)}{(a+b+c-c)^2}\;\text{are in A.P}$

$\text{We know that,}\,\boxed{s=\frac{a+b+c}{2}\implies\,2s=a+b+c}$

$\implies\dfrac{(2c-2s)(2s)}{(2s-c)^2},\dfrac{(2a-2s)(2s)}{(2s-a)^2},\dfrac{(2b-)(2s)}{(2s-b)^2}\;\text{are in A.P}$

$\implies\dfrac{4s(c-s)}{(2s-c)^2},\dfrac{4s(a-s)}{(2s-a)^2},\dfrac{4s(b-s)}{(2s-b)^2}\;\text{are in A.P}$

$\implies\dfrac{-4s(s-c)}{(2s-c)^2},\dfrac{-4s(s-a)}{(2s-a)^2},\dfrac{-4s(s-b)}{(2s-b)^2}\;\text{are in A.P}$

$\text{Divide each term by -4s, we get}$

$\implies\bf\dfrac{s-c}{(2s-c)^2},\dfrac{s-a}{(2s-a)^2},\dfrac{s-b}{(2s-b)^2}\;\text{are in A.P}$

$\text{Hence proved}$