Question

If $1,\frac{x}{2},y$ are in harmonic progression $(x,y\ne 0)$, then the number of integral ordered pairs $(x,y)$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is D $5$

If $1,\frac{x}{2},y$ are in $H.P.$, then $1,\frac{2}{x},\frac{1}{y}$ are in $A.P$

$\frac{2}{x}=\frac{1+\frac{1}{y}}{2}\Rightarrow \frac{4}{x}-\frac{1}{y}=1$ ----- $\left(1\right)$

$4y-x=xy$

$4y-xy+4-x=4$

$(4-x)(y+1)=4$

$x=4-\frac{4}{d},y=d-1$

As $x,y\ne 0\Rightarrow d\ne 1$

No. of division of $4(=2^2)=2+1=3(1,2,4)$

Considering $(-1,-2,-4)$, total no. of factors $=2\times 3=6$

As $d\ne 1$, No of integral pairs $=6-1=5$

$\therefore$ Option $D$ is correct.