Question:

Step 1: Use the AP condition: $2|x+1| = x + |x-1|$. Step 2: Case 1: $x \geq 1$. Then $|x+1| = x+1$, $|x-1| = x-1$. $$2(x+1) = x + (x-1) \Rightarrow 2x+2 = 2x-1 \Rightarrow 2 = -1$$ No solution. Step 3: Case 2: $-1 \leq x < 1$. Then $|x+1| = x+1$, $|x-1| = 1-x$. $$2(x+1) = x + (1-x) \Rightarrow 2x+2 = 1 \Rightarrow x = -\frac{1}{2}$$ First term $a = -1/2$, second term $= |-1/2+1| = 1/2$, common difference $d = 1/2 - (-1/2) = 1$. $$S_{20} = \frac{20}{2}\left[2\left(-\frac{1}{2}\right) + 19(1)\right] = 10[-1+19] = 10 \times 18 = 180$$ Step 4: Case 3: $x < -1$. Then $|x+1| = -(x+1)$, $|x-1| = 1-x$. $$2(-(x+1)) = x + (1-x) \Rightarrow -2x-2 = 1 \Rightarrow x = -\frac{3}{2}$$ First term $a = -3/2$, second term $= |-3/2+1| = 1/2$, $d = 1/2 - (-3/2) = 2$. $$S_{20} = \frac{20}{2}\left[2\left(-\frac{3}{2}\right) + 19(2)\right] = 10[-3+38] = 10 \times 35 = 350$$ Step 5: The possible sums are 180 and 350, corresponding to options (a) and (b).