Question

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 values of $x$ is:

• A. 26
• B. 32
• C. 34
• D. 38
• E. 40

Hint:

When dealing with mean–median–mode problems, always consider the placement of the unknown in the sorted order. Break into cases to find the median. Once identified, enforce the condition (here, AP) to solve for possible values.

Answer 1

Answer: (E)

Step 1: Write down the numbers.
We are given 6 integers: \(20, 4, 10, 4, 8, 4\) and an unknown \(x\). So the set is: \[ \{4, 4, 4, 8, 10, 20, x\} \] 

Step 2: Mode of the set.
Since 4 occurs three times, the mode is always 4 (irrespective of \(x\)). 

Step 3: Mean of the set.
Sum = \(4+4+4+8+10+20+x = 50+x\). Mean = \(\dfrac{50+x}{7}\). 

Step 4: Median of the set.
The median depends on where \(x\) is placed in the sorted order. - If \(x < 4\): order \(x, 4, 4, 4, 8, 10, 20\). Median = 4. - If \(4 < x < 8\): order \(4, 4, 4, x, 8, 10, 20\). Median = \(x\). - If \(x \geq 8\): order \(4, 4, 4, 8, 10, 20, x\). Median = 8. So: \[ \text{Median} = \begin{cases} 4, & x < 4 \\ x, & 4 < x < 8 \\ 8, & x \geq 8 \end{cases} \] 

Step 5: AP condition.
We need {mean, median, mode} in arithmetic progression. Mode = 4. Case 1: \(x < 4\)
Median = 4, Mode = 4. Then AP requires mean = 4. \[ \frac{50+x}{7} = 4 \;\;\Rightarrow\;\; 50+x = 28 \;\;\Rightarrow\;\; x=-22 \] Invalid. Discard. 
Case 2: \(4 < x < 8\)
Median = \(x\), Mode = 4. AP terms: \(\{4, x, \tfrac{50+x}{7}\}\). For AP: \[ x-4 = \frac{50+x}{7} - x \] \[ \Rightarrow 7(x-4) = 50+x-7x \] \[ \Rightarrow 7x-28 = 50-6x \] \[ \Rightarrow 13x = 78 \;\;\Rightarrow\;\; x=6 \] Valid since \(4 < x < 8\). 
Case 3: \(x \geq 8\)
Median = 8, Mode = 4. AP: \(\{4, 8, \tfrac{50+x}{7}\}\). \[ 8-4 = \frac{50+x}{7} - 8 \] \[ 4 = \frac{50+x}{7} - 8 \] \[ \frac{50+x}{7} = 12 \;\;\Rightarrow\;\; 50+x=84 \;\;\Rightarrow\;\; x=34 \] Valid since \(x \geq 8\). 

Step 6: Possible values of \(x\).
So, \(x\) can be \(6\) or \(34\). 

Step 7: Required sum.
\[ 6+34=40 \] 

Final Answer:
\[ \boxed{40} \]