If `x,|x+1|,|x-1|` are first three terms of an A.P., then the sum of its first 20 terms is

Options

A. 360, 180
B. 180,350
C. 150, 100
D. 180, 150

Answer

To solve the problem, we need to determine the values of \( x \) for which the terms \( x, |x+1|, |x-1| \) are in an arithmetic progression (A.P.). Then we will calculate the sum of the first 20 terms of the A.P. based on the values of \( x \) we find.

### 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, this translates to:
\[ 2|x+1| = x + |x-1| \]

### Step 2: Consider cases for \( x \)
We need to analyze the absolute values by considering different cases based on the value of \( x \).

#### Case 1: \( x < -1 \)
In this range:
- \( |x+1| = -(x+1) = -x - 1 \)
- \( |x-1| = -(x-1) = -x + 1 \)

Substituting these into the A.P. condition:
\[ 2(-x - 1) = x + (-x + 1) \]
This simplifies to:
\[ -2x - 2 = 1 \]
\[ -2x = 3 \quad \Rightarrow \quad x = -\frac{3}{2} \]

#### Case 2: \( -1 \leq x < 1 \)
In this range:
- \( |x+1| = x + 1 \)
- \( |x-1| = -(x-1) = -x + 1 \)

Substituting these into the A.P. condition:
\[ 2(x + 1) = x + (-x + 1) \]
This simplifies to:
\[ 2x + 2 = 1 \]
\[ 2x = -1 \quad \Rightarrow \quad x = -\frac{1}{2} \]

#### Case 3: \( x \geq 1 \)
In this range:
- \( |x+1| = x + 1 \)
- \( |x-1| = x - 1 \)

Substituting these into the A.P. condition:
\[ 2(x + 1) = x + (x - 1) \]
This simplifies to:
\[ 2x + 2 = 2x - 1 \]
This leads to a contradiction:
\[ 2 = -1 \quad \text{(not possible)} \]

### Step 3: Identify valid values of \( x \)
The valid values of \( x \) are:
1. \( x = -\frac{3}{2} \)
2. \( x = -\frac{1}{2} \)

### Step 4: Find the A.P. for both values of \( x \)

#### For \( x = -\frac{3}{2} \):
- Terms: \( -\frac{3}{2}, |-\frac{3}{2}+1|, |-\frac{3}{2}-1| \)
- This gives: \( -\frac{3}{2}, \frac{-1}{2}, \frac{5}{2} \)

#### For \( x = -\frac{1}{2} \):
- Terms: \( -\frac{1}{2}, |-\frac{1}{2}+1|, |-\frac{1}{2}-1| \)
- This gives: \( -\frac{1}{2}, \frac{1}{2}, \frac{3}{2} \)

### Step 5: Calculate the common difference and sum of the first 20 terms

#### For \( x = -\frac{3}{2} \):
- First term \( a = -\frac{3}{2} \)
- Common difference \( d = \frac{-1}{2} - \left(-\frac{3}{2}\right) = 1 \)
- Sum of first 20 terms:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
\[ S_{20} = \frac{20}{2} \left(2 \cdot -\frac{3}{2} + 19 \cdot 1\right) = 10 \left(-3 + 19\right) = 10 \cdot 16 = 160 \]

#### For \( x = -\frac{1}{2} \):
- First term \( a = -\frac{1}{2} \)
- Common difference \( d = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1 \)
- Sum of first 20 terms:
\[ S_{20} = \frac{20}{2} \left(2 \cdot -\frac{1}{2} + 19 \cdot 1\right) = 10 \left(-1 + 19\right) = 10 \cdot 18 = 180 \]

### Final Answer
The sum of the first 20 terms can be either 160 or 180 depending on the value of \( x \).