Question

If a, b, c be in G.P. Whereas b - c, c - a, a - b are in H.P., then prove that a + b + c = -3 $\sqrt{\left(ac\right)}$

Answer 1

$b^2=ac$
$\frac{2}{c-a}=\frac{1}{b-c}+\frac{1}{a-b}=\frac{a-c}{(b-c)(a-b)}$
or $2(ab-ca-b^2+bc)=-(a-c{)}^2$
or $2(ab-2b^2+bc)=-(a-c{)}^2$
or 2b (a + c - 2 $\sqrt{ac}$) = -$\left[\right(\sqrt{a}-\sqrt{c}\left)\right(\sqrt{a}+\sqrt{c}){]}^2$
or $2b(\sqrt{a}-\sqrt{c}{)}^2=(\sqrt{a}-\sqrt{c}{)}^2(\sqrt{a}+\sqrt{c}{)}^2$
$\therefore 2b=-(a+c+2\sqrt{ac})asa\ne c$
$\therefore a+c=-4b,by\left(1\right)$
or a + b + c = -3b = $-3\sqrt{ac}$ by(1)