Question

If $a,b,c$ are in $G.P$ and $b-c,c-a,a-b$ are in $H.P$, then prove that $a+b+c=3(ac{)}^{1/2}$

Answer 1

$Givena,b,careinGP\therefore b^2=acb-c,c-a,a-binHP\left(\frac{1}{b-c}\right),\left(\frac{1}{c-a}\right),\left(\frac{1}{a-b}\right)inAPor\left(\frac{2}{c-a}\right)=\left(\frac{1}{b-c}\right)+\left(\frac{1}{a-b}\right)\left(\frac{2}{c-a}\right)=\left(\frac{a-b+b-c}{(b-c)(a-b)}\right)or(c-a)(a-c)=2(b-c)(a-b)-(a-c{)}^2=2(b-c\left)\right(a-b)a^2+c^2-2ac=-2(ba-b^2-ca+bc)a^2+c^2-2ac=-2ba+2b^2+2ca-2bcor2ba+2bc=2b^2-a^2-c^2+4ac\therefore (a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2+2b^2-a^2-c^2+4ac+2ca=3b^2+6ac=3b^2+6b^2=9b^2\therefore a+b+c=3b=3(ac{)}^{1/2}$