# Let x be the A.M. and y, z be two G.M.s between two positive numbers.

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)]

$=2$