To solve the problem of finding the sum \( a + b + c + d + e + f \) where \( a, b, c, d, e, f \) are arithmetic means between 2 and 12, we can follow these steps:
### Step 1: Understand the Problem
We need to insert 6 arithmetic means between the numbers 2 and 12. This means we will have a total of 8 terms in the sequence: \( 2, a, b, c, d, e, f, 12 \).
### Step 2: Identify the Common Difference
In an arithmetic sequence, the difference between consecutive terms is constant. Let the common difference be \( d \).
The first term \( a_1 = 2 \) and the last term \( a_8 = 12 \). We can express the \( n \)-th term of an arithmetic sequence as:
\[
a_n = a_1 + (n-1)d
\]
For the 8th term:
\[
a_8 = 2 + 7d = 12
\]
### Step 3: Solve for the Common Difference
Now, we can solve for \( d \):
\[
2 + 7d = 12
\]
Subtract 2 from both sides:
\[
7d = 10
\]
Now, divide by 7:
\[
d = \frac{10}{7}
\]
### Step 4: Calculate Each Arithmetic Mean
Now we can find each term \( a, b, c, d, e, f \):
- \( a = 2 + d = 2 + \frac{10}{7} = \frac{14}{7} + \frac{10}{7} = \frac{24}{7} \)
- \( b = 2 + 2d = 2 + 2 \cdot \frac{10}{7} = 2 + \frac{20}{7} = \frac{14}{7} + \frac{20}{7} = \frac{34}{7} \)
- \( c = 2 + 3d = 2 + 3 \cdot \frac{10}{7} = 2 + \frac{30}{7} = \frac{14}{7} + \frac{30}{7} = \frac{44}{7} \)
- \( d = 2 + 4d = 2 + 4 \cdot \frac{10}{7} = 2 + \frac{40}{7} = \frac{14}{7} + \frac{40}{7} = \frac{54}{7} \)
- \( e = 2 + 5d = 2 + 5 \cdot \frac{10}{7} = 2 + \frac{50}{7} = \frac{14}{7} + \frac{50}{7} = \frac{64}{7} \)
- \( f = 2 + 6d = 2 + 6 \cdot \frac{10}{7} = 2 + \frac{60}{7} = \frac{14}{7} + \frac{60}{7} = \frac{74}{7} \)
### Step 5: Calculate the Sum
Now we can find the sum:
\[
a + b + c + d + e + f = \frac{24}{7} + \frac{34}{7} + \frac{44}{7} + \frac{54}{7} + \frac{64}{7} + \frac{74}{7}
\]
Combine the fractions:
\[
= \frac{24 + 34 + 44 + 54 + 64 + 74}{7} = \frac{294}{7} = 42
\]
### Final Answer
Thus, the sum \( a + b + c + d + e + f = 42 \).
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