Question

If $A,G,H$ be respectively A.M., GM., H.M. of three numbers (>0) then the equation whose roots are the numbers is given by

• A. $x^3+3A{x}^2+3\frac{G}{H}x-G^3=0$
• B. $x^3-3A{x}^2-3\left(\frac{G^3}{H}\right)+G^3=0$
• C. $x^3-3A{x}^2+G^3(3x-1)=0$
• D. $x^3-3A{x}^2+3\left(\frac{G^3}{H}\right)x-G^3=0$

Answer 1

The correct option is D $x^3-3A{x}^2+3\left(\frac{G^3}{H}\right)x-G^3=0$

Let A,G,H be respectively A.M., GM., H.M. of three numbers $a,b,c$
Therefore, $A=\frac{a+b+c}{3}$
$\Rightarrow a+b+c=3A$ ....(1)
and $G={\left(abc\right)}^{\frac{1}{3}}$
$\Rightarrow abc=G^3$ ....(2)

and $H=\frac{3abc}{ab+bc+ac}$
$\Rightarrow ab+bc+ca=\frac{3abc}{H}=\frac{3G^3}{H}$ ....(3)

Cubic equation with roots $a,b,c$ can be written as
$x^3-(a+b+c)x^2+(ab+bc+ac)x-abc=0$
Therefore, $x^3-3A{x}^2+3\left(\frac{G^3}{H}\right)x-G^3=0$ ....[ From (1), (2) & (3)]
Ans :D