There are n A.M.'s between 3 and 29 such that 6th mean : (n - 1)th mean : : 3 : 5 then the value of n is

Options

A. 10
B. 11
C. 12
D. none

Answer

To solve the problem, we need to find the value of \( n \) such that the ratio of the 6th arithmetic mean (A.M.) to the \( (n-1) \)th arithmetic mean is \( 3:5 \). The two endpoints of the A.M. sequence are 3 and 29.

### Step-by-Step Solution:

1. **Identify the first term and the last term:**
- The first term \( a = 3 \)
- The last term \( l = 29 \)

2. **Determine the number of terms:**
- The total number of terms in the sequence is \( n + 2 \) (including the two endpoints).
- Therefore, the number of A.M.s is \( n \).

3. **Calculate the common difference \( d \):**
- The common difference \( d \) can be calculated using the formula:
\[ d = \frac{l - a}{n + 1} = \frac{29 - 3}{n + 1} = \frac{26}{n + 1} \]

4. **Find the 6th A.M. and the \( (n-1) \)th A.M.:**
- The 6th A.M. is given by:
\[ A_6 = a + 6d = 3 + 6 \left(\frac{26}{n + 1}\right) = 3 + \frac{156}{n + 1} \]
- The \( (n-1) \)th A.M. is given by:
\[ A_{n-1} = a + (n-1)d = 3 + (n-1) \left(\frac{26}{n + 1}\right) = 3 + \frac{26(n-1)}{n + 1} \]

5. **Set up the ratio:**
- According to the problem, the ratio of the 6th A.M. to the \( (n-1) \)th A.M. is:
\[ \frac{A_6}{A_{n-1}} = \frac{3}{5} \]
- Substituting the expressions for \( A_6 \) and \( A_{n-1} \):
\[ \frac{3 + \frac{156}{n + 1}}{3 + \frac{26(n-1)}{n + 1}} = \frac{3}{5} \]

6. **Cross-multiply to eliminate the fraction:**
- Cross-multiplying gives:
\[ 5 \left(3 + \frac{156}{n + 1}\right) = 3 \left(3 + \frac{26(n-1)}{n + 1}\right) \]

7. **Simplify the equation:**
- Expanding both sides:
\[ 15 + \frac{780}{n + 1} = 9 + \frac{78(n - 1)}{n + 1} \]
- Rearranging gives:
\[ 15 - 9 = \frac{78(n - 1)}{n + 1} - \frac{780}{n + 1} \]
- This simplifies to:
\[ 6(n + 1) = 78(n - 1) - 780 \]

8. **Solve for \( n \):**
- Rearranging gives:
\[ 6n + 6 = 78n - 78 - 780 \]
\[ 6n + 6 = 78n - 858 \]
\[ 858 + 6 = 78n - 6n \]
\[ 864 = 72n \]
\[ n = \frac{864}{72} = 12 \]

### Final Answer:
The value of \( n \) is \( 12 \).