If the lines x+2ay+a=0, x+3by+b=0, x+4cy+c=0 are concurrent, then a,b,c are in

$\begin{array}{l}Givenlines\\ x+2ay+a=0\\ x+3by+b=0\\ x+4cy+c=0\\ Thelinesareconcurrentiff\\ \left|\begin{array}{ccc}1& 2a& a\\ 1& 3b& b\\ 1& 4c& c\end{array}\right|=0\\ \\ \Rightarrow 1(3bc-4bc)-2a(c-b)+a(4c-3b)=0\\ \Rightarrow -bc-2ac+2ab+4ac-3ab=0\\ \Rightarrow -ab-bc+2ac=0\\ \Rightarrow 2ac=ab+bc\\ Dividingbyabc\\ \Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\\ a,b,careinH.P\end{array}$