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

Answer

To solve the problem, we need to find the value of \( n \) given that there are \( n \) arithmetic means (A.M.s) between 3 and 29, and the ratio of the 6th mean to the \( (n-1) \)th mean is \( 3:5 \).

### Step-by-Step Solution:

1. **Identify the Arithmetic Means**:
The arithmetic means between 3 and 29 can be represented as:
\[ 3, a_1, a_2, \ldots, a_n, 29 \]
where \( a_i = 3 + id \) for \( i = 1, 2, \ldots, n \) and \( d \) is the common difference.

2. **Find the Common Difference**:
Since 29 is the last term after \( a_n \), we can write:
\[ 29 = 3 + (n + 1)d \]
Simplifying this gives:
\[ 26 = (n + 1)d \quad \Rightarrow \quad d = \frac{26}{n + 1} \quad \text{(Equation 1)} \]

3. **Express the 6th and \( (n-1) \)th Means**:
The 6th mean \( a_6 \) is:
\[ a_6 = 3 + 6d \]
The \( (n-1) \)th mean \( a_{n-1} \) is:
\[ a_{n-1} = 3 + (n-1)d \]

4. **Set Up the Ratio**:
According to the problem, we have:
\[ \frac{a_6}{a_{n-1}} = \frac{3}{5} \]
Substituting the expressions for \( a_6 \) and \( a_{n-1} \):
\[ \frac{3 + 6d}{3 + (n-1)d} = \frac{3}{5} \]

5. **Cross-Multiply**:
Cross-multiplying gives:
\[ 5(3 + 6d) = 3(3 + (n-1)d) \]
Expanding both sides:
\[ 15 + 30d = 9 + 3(n-1)d \]

6. **Simplify the Equation**:
Rearranging gives:
\[ 15 + 30d - 9 = 3(n-1)d \]
\[ 6 + 30d = 3(n-1)d \]

7. **Factor Out \( d \)**:
Rearranging the equation:
\[ 6 = 3(n-1)d - 30d \]
\[ 6 = (3n - 3 - 30)d \]
\[ 6 = (3n - 33)d \]
Thus:
\[ d = \frac{6}{3n - 33} \quad \text{(Equation 2)} \]

8. **Equate the Two Expressions for \( d \)**:
From Equation 1 and Equation 2:
\[ \frac{26}{n + 1} = \frac{6}{3n - 33} \]

9. **Cross-Multiply Again**:
Cross-multiplying gives:
\[ 26(3n - 33) = 6(n + 1) \]
Expanding both sides:
\[ 78n - 858 = 6n + 6 \]

10. **Rearranging to Solve for \( n \)**:
Bringing all terms involving \( n \) to one side:
\[ 78n - 6n = 858 + 6 \]
\[ 72n = 864 \]
Dividing both sides by 72:
\[ n = 12 \]

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