$2\left|x+1\right|=x+\left|x-1\right|$
(i) Case- $\mathrm{{\rm I}}$ :-

$x\le -1$
$2\left(-x-1\right)=x+\left(-x+1\right)\Rightarrow -2x-2=1$
$\Rightarrow x=-\frac{3}{2}$
Then the A.P. is $\to -\frac{3}{2},\frac{1}{2},\frac{5}{2}.\mathrm{}\mathrm{}.\mathrm{}\mathrm{}.\mathrm{}\mathrm{}.\mathrm{}\mathrm{}.\mathrm{}\mathrm{}.$
$S_{20}=\frac{20}{2}\left[-3+19\times 2\right]=\frac{20}{2}\times 35=350$
(ii) Case- $\mathrm{{\rm I}}\mathrm{{\rm I}}$ :-
$-1\le x<1$
$2\left(x+1\right)=x+\left(-x+1\right)\Rightarrow 2x+2=1\Rightarrow x=-\frac{1}{2}$
Then the A.P. is $\to -\frac{1}{2},\frac{1}{2},\frac{3}{2}.\mathrm{}\mathrm{}.\mathrm{}\mathrm{}.\mathrm{}\mathrm{}.\mathrm{}\mathrm{}.\mathrm{}\mathrm{}.$
$S_{20}=\frac{20}{2}\left[-1+19\times 1\right]=10\times 18=180$
(ii) Case- $\mathrm{{\rm I}}\mathrm{{\rm I}}\mathrm{{\rm I}}$ :-
$x\ge 1$
$2\left(x+1\right)=x+x-1\Rightarrow \cancel{2x}+2=\cancel{2x}-1\Rightarrow 2=-1$
(not possible)
Hence, the required sum is $350$ or $180$