Question:
Let $x$ be the A.M. and $y, z$ be two G.M.s between two positive numbers. Then, $\frac{y^{3}+z^{3}}{x y z}$ is equal to
(a) 1
(b) 2
(c) $\frac{1}{2}$
(d) none of these
Solution:
(b) 2
Let the two numbers be $a$ and $b$.
$a, x$ and $b$ are in A.P.
$\therefore 2 x=a+b$ ...(i)
Also, $a, y, z$ and $b$ are in G.P.
$\therefore \frac{y}{a}=\frac{z}{y}=\frac{b}{z}$
$\Rightarrow y^{2}=a z, y z=a b, z^{2}=b y$ ...(ii)
Now, $\frac{y^{3}+z^{3}}{x y z}$
$=\frac{y^{2}}{x z}+\frac{z^{2}}{x y}$
$=\frac{1}{x}\left(\frac{y^{2}}{z}+\frac{z^{2}}{y}\right)$
$=\frac{1}{x}\left(\frac{a z}{z}+\frac{b y}{y}\right)$ [Using (ii)]
$=\frac{1}{x}(a+b)$
$=\frac{2}{(a+b)}(a+b)$ [Using (i)]
$=2$