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 $H.P.$
$\frac{2}{x}=\frac{1}{4}+\frac{1}{y}$
$\Rightarrow \frac{2}{x}-\frac{1}{4}=\frac{1}{y}$
$\Rightarrow \frac{8-x}{4x}=\frac{1}{y}$
$\Rightarrow y=\frac{4x}{8-x}=\frac{4(8-(8-x))}{8-x}=\frac{32}{8-x}-4$
$8-x$ 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 $3$.