Suppose that  a,b,c are in arithmetic progression then the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$  passes through the fixed point $\left(\mathrm{l},\mathrm{m}\right)$ then $\mathrm{l}+\mathrm{m}=$

Given a,b,c  are in arithmetic progression, so that  $\mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2}$
Substitute $\mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2}$  in the equation   $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$
$\begin{array}{c}\mathrm{ax}+\left(\frac{\mathrm{a}+\mathrm{c}}{2}\right)\mathrm{y}+\mathrm{c}=0\\ 2\mathrm{ax}+\left(\mathrm{a}+\mathrm{c}\right)\mathrm{y}+2\mathrm{c}=0\\ \mathrm{a}\left(2\mathrm{x}+\mathrm{y}\right)+\mathrm{c}\left(\mathrm{y}+2\right)=0\end{array}$  
This represents the set of all lines passing through the point of intersection of $2\mathrm{x}+\mathrm{y}=0$  and  $\mathrm{y}+2=0$
The point of intersection is  $\left(1,-2\right)$
Hence $\mathrm{l}+\mathrm{m}=1-2=\boxed{-1}$