`{:("Column A"," ","Column B"),("The average of five consecutive integers starting from m",,"The average of six consecutive integers staring from m"):}`

Options

A. If column A is larger
B. If column B is larger
C. If the columns are equal
D. If there is not enough information to decide

Answer

To solve the problem, we need to find the averages of two sets of consecutive integers starting from \( m \): one set consists of five integers and the other consists of six integers. We will then compare the two averages.

### Step 1: Calculate the average of five consecutive integers starting from \( m \).

The five consecutive integers starting from \( m \) are:
- \( m \)
- \( m + 1 \)
- \( m + 2 \)
- \( m + 3 \)
- \( m + 4 \)

To find the average, we sum these integers and divide by 5:
\[ \text{Average}_5 = \frac{m + (m + 1) + (m + 2) + (m + 3) + (m + 4)}{5} \]
Calculating the sum:
\[ = \frac{5m + (1 + 2 + 3 + 4)}{5} = \frac{5m + 10}{5} \]
\[ = m + 2 \]

### Step 2: Calculate the average of six consecutive integers starting from \( m \).

The six consecutive integers starting from \( m \) are:
- \( m \)
- \( m + 1 \)
- \( m + 2 \)
- \( m + 3 \)
- \( m + 4 \)
- \( m + 5 \)

To find the average, we sum these integers and divide by 6:
\[ \text{Average}_6 = \frac{m + (m + 1) + (m + 2) + (m + 3) + (m + 4) + (m + 5)}{6} \]
Calculating the sum:
\[ = \frac{6m + (1 + 2 + 3 + 4 + 5)}{6} = \frac{6m + 15}{6} \]
\[ = m + 2.5 \]

### Step 3: Compare the two averages.

Now we have:
- Average of five consecutive integers: \( m + 2 \)
- Average of six consecutive integers: \( m + 2.5 \)

To compare:
\[ m + 2.5 > m + 2 \]

This shows that the average of six consecutive integers is greater than the average of five consecutive integers.

### Conclusion:

Since \( m + 2.5 \) is greater than \( m + 2 \), we conclude that the value in **Column B** is larger than the value in **Column A**.

**Final Answer:** Column B is larger.

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