Mod AP

For the terms to be in AP, $2\left| x+1 \right| =x+\left| x-1 \right|$ Case I: if $x\geq1$ : $2x+2=x+x-1 \Rightarrow 2 = -1$ So, no solution in $(1,\infty)$

Case II: if $-1 \leq x <1): \[ 2x+2=x-x+1 \Rightarrow x = -\frac{1}{2}$

Case III: if $x<-1$ : $-2x-2=x-x+1 \Rightarrow x = -\frac{3}{2}$

Solving this equation, we get two solutions, $-\frac{1}{2}$ and $-\frac{3}{2}$

Thus, first three terms of AP are, $-\frac{1}{2}\ , \frac{1}{2}\ and \frac{3}{2}$ or, $-\frac{3}{2}\, \frac{1}{2}\, \frac{5}{2}\$

Common difference = $d = 1 or 2$ for each case respectively.

So, $\sum_{i=1}^{20}a+(i-1)d = \frac{20 \times \left(-\frac{1}{2} \times 2 + (20 -1)\times 1 \right) }{2} = \boxed{180}$

Or, $\sum_{i=1}^{20}a+(i-1)d = \frac{20 \times \left(-\frac{3}{2} \times 2 + (20 -1)\times 2 \right) }{2} = \boxed{350}$