If a, b, c, d, e, f are A.M's between 2 and 12, then a + b + c + d + e + f is equal to

Options

A. 14
B. 42
C. 84
D. none of these

Answer

To solve the problem of finding the sum of the arithmetic means (A.M.s) between 2 and 12, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Terms**:
We need to find 6 A.M.s between 2 and 12. This means we will have a total of 8 terms: the first term (2), the 6 A.M.s (let's call them a, b, c, d, e, f), and the last term (12).

2. **Determine the Common Difference**:
The formula for the nth term of an arithmetic progression (AP) is given by:
\[ T_n = a + (n-1)d \]
Here, \(T_1 = 2\) (the first term), \(T_8 = 12\) (the eighth term), and \(n = 8\).
We can set up the equation:
\[ 12 = 2 + (8-1)d \]
Simplifying this gives:
\[ 12 = 2 + 7d \]
\[ 10 = 7d \implies d = \frac{10}{7} \]

3. **Write the A.M.s**:
Now we can express each A.M. in terms of \(d\):
- \(a = 2 + d = 2 + \frac{10}{7} = \frac{24}{7}\)
- \(b = 2 + 2d = 2 + 2 \times \frac{10}{7} = \frac{34}{7}\)
- \(c = 2 + 3d = 2 + 3 \times \frac{10}{7} = \frac{44}{7}\)
- \(d = 2 + 4d = 2 + 4 \times \frac{10}{7} = \frac{54}{7}\)
- \(e = 2 + 5d = 2 + 5 \times \frac{10}{7} = \frac{64}{7}\)
- \(f = 2 + 6d = 2 + 6 \times \frac{10}{7} = \frac{74}{7}\)

4. **Sum the A.M.s**:
Now we can sum all the A.M.s:
\[ a + b + c + d + e + f = \left(\frac{24}{7} + \frac{34}{7} + \frac{44}{7} + \frac{54}{7} + \frac{64}{7} + \frac{74}{7}\right) \]
This can be simplified as:
\[ = \frac{24 + 34 + 44 + 54 + 64 + 74}{7} = \frac{294}{7} = 42 \]

### Final Answer:
Thus, the sum \(a + b + c + d + e + f\) is equal to **42**.

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