If three non-zero real numbers a,b,c are in Harmonic progression then the straight lines $\frac{x}{a}+\frac{y}{b}-\frac{2}{c}=0$ are concurrent at the point

Given line $\frac{x}{a}+\frac{y}{b}-\frac{2}{c}=0---\left(i\right)$

and given a,b,c are in Harmonic progression 

$\begin{array}{l}\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\\ \Rightarrow \frac{1}{c}=\frac{2}{b}-\frac{1}{a}\end{array}$

sub in (i)

$\begin{array}{l}\frac{x}{a}+\frac{y}{b}-2\left(\frac{2}{b}-\frac{1}{a}\right)=0\\ \Rightarrow \frac{x}{a}+\frac{y}{b}-\frac{4}{b}+\frac{2}{a}=0\\ \Rightarrow \frac{1}{a}\left(x+2\right)+\frac{1}{b}\left(y-4\right)=0\\ \Rightarrow \left(x+2\right)+\frac{a}{b}\left(y-4\right)=0\end{array}$

If is of the form $L_1+\lambda L_2=0$

$\begin{array}{l} Where L_1=x+2=0\Rightarrow x=-2\\ L_2=y-4=0\Rightarrow y=4\end{array}$

$\therefore$ Point of intersection is (-2,4)