There are `n AM's` between `1 & 31` such that `7th` mean `: (n-1)th` mean`= 5:9`, then find the value of` n`

Answer

To solve the problem, we need to find the number of arithmetic means (AMs) between 1 and 31, given that the ratio of the 7th mean to the (n-1)th mean is 5:9.

### Step-by-step Solution:

1. **Understanding the Arithmetic Progression (AP)**:
- The first term (a) is 1.
- The last term (l) is 31.
- There are `n` arithmetic means inserted between 1 and 31, making the total number of terms in the AP equal to `n + 2` (including 1 and 31).

2. **Finding the Common Difference (d)**:
- The formula for the nth term of an AP is given by:
\[ l = a + (n + 1) \cdot d \]
- Substituting the known values:
\[ 31 = 1 + (n + 1) \cdot d \]
- Rearranging gives:
\[ (n + 1) \cdot d = 31 - 1 = 30 \]
- Thus, we have:
\[ d = \frac{30}{n + 1} \]

3. **Finding the 7th and (n-1)th Means**:
- The 7th mean (a7) can be expressed as:
\[ a_7 = a + 7d = 1 + 7d \]
- The (n-1)th mean (a(n-1)) can be expressed as:
\[ a_{n-1} = a + (n - 1)d = 1 + (n - 1)d \]

4. **Setting up the Ratio**:
- We are given that:
\[ \frac{a_7}{a_{n-1}} = \frac{5}{9} \]
- Substituting the expressions for a7 and a(n-1):
\[ \frac{1 + 7d}{1 + (n - 1)d} = \frac{5}{9} \]

5. **Cross Multiplying**:
- Cross-multiplying gives:
\[ 9(1 + 7d) = 5(1 + (n - 1)d) \]
- Expanding both sides:
\[ 9 + 63d = 5 + 5(n - 1)d \]
- This simplifies to:
\[ 9 + 63d = 5 + 5nd - 5d \]
- Rearranging gives:
\[ 63d + 5d - 5nd = 5 - 9 \]
- Which simplifies to:
\[ 68d - 5nd = -4 \]

6. **Substituting d**:
- Substitute \( d = \frac{30}{n + 1} \) into the equation:
\[ 68 \left(\frac{30}{n + 1}\right) - 5n \left(\frac{30}{n + 1}\right) = -4 \]
- Multiplying through by \( n + 1 \) to eliminate the denominator:
\[ 68 \cdot 30 - 5n \cdot 30 = -4(n + 1) \]
- This simplifies to:
\[ 2040 - 150n = -4n - 4 \]
- Rearranging gives:
\[ 2040 + 4 = 150n - 4n \]
\[ 2044 = 146n \]
- Thus:
\[ n = \frac{2044}{146} = 14 \]

### Final Answer:
The value of \( n \) is **14**.