To solve the problem where \( x, |x + 1|, |x - 1| \) are three terms of an arithmetic progression (A.P.), we will follow these steps:
### Step 1: Set up the condition for A.P.
For three terms \( a, b, c \) to be in A.P., the condition is:
\[
2b = a + c
\]
In our case, we have:
\[
a = x, \quad b = |x + 1|, \quad c = |x - 1|
\]
Thus, we can write:
\[
2|x + 1| = x + |x - 1|
\]
### Step 2: Analyze cases based on the value of \( x \)
Since the absolute values can change based on the value of \( x \), we will consider three cases:
1. \( x > 1 \)
2. \( -1 \leq x \leq 1 \)
3. \( x < -1 \)
### Case 1: \( x > 1 \)
In this case, both \( |x + 1| \) and \( |x - 1| \) are positive, so:
\[
2(x + 1) = x + (x - 1)
\]
This simplifies to:
\[
2x + 2 = 2x - 1
\]
This results in:
\[
2 = -1 \quad \text{(no solution)}
\]
### Case 2: \( -1 \leq x \leq 1 \)
Here, \( |x + 1| = x + 1 \) and \( |x - 1| = 1 - x \):
\[
2(x + 1) = x + (1 - x)
\]
This simplifies to:
\[
2x + 2 = 1
\]
Solving for \( x \):
\[
2x = -1 \quad \Rightarrow \quad x = -\frac{1}{2}
\]
### Step 3: Find the terms of the A.P.
Substituting \( x = -\frac{1}{2} \):
\[
a = -\frac{1}{2}, \quad b = |-\frac{1}{2} + 1| = \frac{1}{2}, \quad c = |-\frac{1}{2} - 1| = \frac{3}{2}
\]
Thus, the terms are:
\[
-\frac{1}{2}, \frac{1}{2}, \frac{3}{2}
\]
### Step 4: Calculate the common difference
The common difference \( d \) is:
\[
d = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1
\]
### Step 5: Find the sum of the first 20 terms
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[
S_n = \frac{n}{2} \left(2a + (n - 1)d\right)
\]
Substituting \( n = 20 \), \( a = -\frac{1}{2} \), and \( d = 1 \):
\[
S_{20} = \frac{20}{2} \left(2 \left(-\frac{1}{2}\right) + (20 - 1) \cdot 1\right)
\]
\[
= 10 \left(-1 + 19\right) = 10 \cdot 18 = 180
\]
### Case 3: \( x < -1 \)
In this case, both \( |x + 1| \) and \( |x - 1| \) are negative:
\[
2(-x - 1) = x - (x - 1)
\]
This simplifies to:
\[
-2x - 2 = 1
\]
Solving for \( x \):
\[
-2x = 3 \quad \Rightarrow \quad x = -\frac{3}{2}
\]
### Step 6: Find the terms of the A.P. for Case 3
Substituting \( x = -\frac{3}{2} \):
\[
a = -\frac{3}{2}, \quad b = |-\frac{3}{2} + 1| = \frac{1}{2}, \quad c = |-\frac{3}{2} - 1| = \frac{5}{2}
\]
Thus, the terms are:
\[
-\frac{3}{2}, \frac{1}{2}, -\frac{5}{2}
\]
### Step 7: Calculate the common difference for Case 3
The common difference \( d \) is:
\[
d = \frac{1}{2} - \left(-\frac{3}{2}\right) = 2
\]
### Step 8: Find the sum of the first 20 terms for Case 3
Using the same formula for \( S_n \):
\[
S_{20} = \frac{20}{2} \left(2 \left(-\frac{3}{2}\right) + (20 - 1) \cdot 2\right)
\]
\[
= 10 \left(-3 + 38\right) = 10 \cdot 35 = 350
\]
### Final Result
The sum of the first 20 terms can either be \( 180 \) or \( 350 \) depending on the value of \( x \).
