Question

The number of ordered pairs $(x, y),$ where $x, y \in N$ for which $4, x, y$ are in $H.P. ,$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

$4, x, y$ are in $H.P.$
$\frac{2}{x}=\frac{1}{4}+\frac{1}{y}$
$\Rightarrow \frac{2}{x}-\frac{1}{4}=\frac{1}{y}$
$\Rightarrow \frac{8-x}{4 x}=\frac{1}{y}$
$\Rightarrow y=\frac{4 x}{8-x}=\frac{4(8-(8-x))}{8-x}=\frac{32}{8-x}-4$
$8-x$ must be a factor of $32$
$8-x=1 \,\,\, \Rightarrow \,\,\, x=7, y=28$
$8-x=2 \,\,\, \Rightarrow \,\,\, x=6, y=12 $
$8-x=4 \,\,\, \Rightarrow \,\,\, x=4, y=4$
$8-x=8\,\,\, \Rightarrow \,\,\, x=0, y=0$ (Not possible)
$\therefore $ Number of ordered pairs of $(x, y)$ is $3 $.