In a list of 7 integers, one integer, denoted as x is unknown. The other six integers are 20, 4, 10, 4, 8 and 4. If the mean, median , and mode of these seven integers are arranged in increasing order, they form an arithmetic progression. The sum of all possible ways of x is

Options

A. 26
B. 32
C. 40
D. 38

Answer

To solve the problem, we need to find the integer \( x \) such that the mean, median, and mode of the seven integers (including \( x \)) form an arithmetic progression (AP). The known integers are 20, 4, 10, 4, 8, and 4.

### Step-by-Step Solution:

1. **List the Known Integers**:
The known integers are 20, 4, 10, 4, 8, and 4.

2. **Sort the Known Integers**:
Arranging these integers in ascending order gives us:
\[ 4, 4, 4, 8, 10, 20 \]

3. **Identify the Mode**:
The mode is the number that appears most frequently. In this case, the mode is:
\[ \text{Mode} = 4 \]

4. **Determine the Median**:
The median depends on the position of \( x \) in the sorted list. Since there are 7 integers, the median will be the 4th number when arranged in order. We will consider different cases for \( x \).

5. **Calculate the Mean**:
The mean is calculated as:
\[ \text{Mean} = \frac{(20 + 4 + 10 + 4 + 8 + 4 + x)}{7} = \frac{50 + x}{7} \]

6. **Case 1: \( x \leq 4 \)**:
- Mode = 4
- Median = 4 (since \( x \) will be among the 4s)
- Mean = \(\frac{50 + x}{7}\)
- The values are \( 4, 4, \frac{50 + x}{7} \). They do not form an AP since the mean cannot equal 4.

7. **Case 2: \( 4 < x < 8 \)**:
- Mode = 4
- Median = \( x \) (since \( x \) will be in the middle)
- Mean = \(\frac{50 + x}{7}\)
- The values are \( 4, x, \frac{50 + x}{7} \). For these to form an AP, we need:
\[ 2x = 4 + \frac{50 + x}{7} \]
Multiplying through by 7:
\[ 14x = 28 + 50 + x \implies 13x = 78 \implies x = 6 \]

8. **Case 3: \( x \geq 8 \)**:
- Mode = 4
- Median = 8 (since \( x \) is greater than 8)
- Mean = \(\frac{50 + x}{7}\)
- The values are \( 4, 8, \frac{50 + x}{7} \). For these to form an AP, we need:
\[ 2 \times 8 = 4 + \frac{50 + x}{7} \]
Multiplying through by 7:
\[ 112 = 28 + 50 + x \implies x = 34 \]

9. **Final Values of \( x \)**:
The possible values of \( x \) that satisfy the conditions are \( 6 \) and \( 34 \).

10. **Sum of All Possible Values of \( x \)**:
\[ 6 + 34 = 40 \]

### Final Answer:
The sum of all possible values of \( x \) is \( \boxed{40} \).