If `n` arithmetic means are inserted between `7` and `71` such that `5^(th)` A.M. is `27` then `n= ?`

Options

A. 10
B. 11
C. 12
D. None of these

Explanation

To solve the problem, we need to find the number of arithmetic means \( n \) inserted between the numbers 7 and 71 such that the 5th arithmetic mean is 27.

### Step-by-Step Solution:

1. **Identify the first and last terms:**
- The first term \( a_1 = 7 \)
- The last term \( a_{n+2} = 71 \)

2. **Determine the number of terms:**
- The total number of terms is \( n + 2 \) (including the first and last terms).

3. **Calculate the common difference \( d \):**
- The formula for the common difference \( d \) in an arithmetic series is given by:
\[ d = \frac{\text{last term} - \text{first term}}{\text{number of terms} - 1} \]
- Substituting the values we have:
\[ d = \frac{71 - 7}{(n + 2) - 1} = \frac{64}{n + 1} \]

4. **Express the 5th term:**
- The 5th term \( a_5 \) can be expressed as:
\[ a_5 = a_1 + 4d \]
- Since \( a_5 = 27 \), we can write:
\[ 27 = 7 + 4d \]

5. **Substitute \( d \) into the equation:**
- From the previous step, we have:
\[ 27 = 7 + 4\left(\frac{64}{n + 1}\right) \]
- Simplifying this:
\[ 27 - 7 = 4\left(\frac{64}{n + 1}\right) \]
\[ 20 = \frac{256}{n + 1} \]

6. **Cross-multiply to solve for \( n + 1 \):**
- Cross-multiplying gives:
\[ 20(n + 1) = 256 \]
- Expanding this:
\[ 20n + 20 = 256 \]
- Rearranging gives:
\[ 20n = 256 - 20 \]
\[ 20n = 236 \]
\[ n = \frac{236}{20} = 11.8 \]

7. **Since \( n \) must be a whole number, we round down:**
- Since \( n \) must be an integer, we take \( n = 11 \).

### Conclusion:
The value of \( n \) is \( 11 \).