Question

If $x$ be the $AM$ and $y,z$ be two $GM$'s between two positive numbers, then $\frac{y^3+z^3}{xyz}$ is equal to -

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

$Letthenumbersbea,bthen$
$a,x,bareinA.P$
$2x=a+b...............\left(i\right)$
$Also$
$a,y,z,bareinG.P$
$\frac{y}{a}=\frac{z}{y}=\frac{b}{z}$
$\Rightarrow y^2=az,yz=ab,z^2=yb.........\left(ii\right)$
$Now$
$\frac{y^3+z^3}{xyz}$
$=\frac{y^2}{xz}+\frac{z^2}{yx}$
$=\frac{1}{x}(\frac{y^2}{z}+\frac{z^2}{y})$
$=\frac{1}{x}(\frac{az}{z}+\frac{by}{y})u\sin g\left(ii\right)$
$=\frac{1}{x}(a+b)$
$=\frac{2}{a+b}(a+b)u\sin g\left(i\right)$
$=2$