There are m A.M.s' between 1 and 31 . If the ratio of the `7^(th)` and `(m-1)^(th)` means is 5:9 then m=

Options

A. 10
B. 12
C. 14
D. 16

Answer

To solve the problem, we need to find the value of \( m \) given that there are \( m \) arithmetic means between 1 and 31, and the ratio of the 7th and (m-1)th means is \( 5:9 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a sequence starting from 1 to 31 with \( m \) arithmetic means inserted. This means there are \( m + 2 \) terms in total (including 1 and 31).

2. **Finding the Common Difference**:
The first term \( a = 1 \) and the last term \( l = 31 \). The total number of terms \( n = m + 2 \).
The formula for the last term of an arithmetic sequence is:
\[ l = a + (n-1)d \]
Substituting the known values:
\[ 31 = 1 + (m + 1)d \]
Simplifying this gives:
\[ 30 = (m + 1)d \quad \text{(Equation 1)} \]

3. **Finding the 7th and (m-1)th Means**:
The 7th mean \( A_7 \) and the (m-1)th mean \( A_{m-1} \) can be expressed as:
\[ A_7 = a + 7d = 1 + 7d \]
\[ A_{m-1} = a + (m-1)d = 1 + (m-1)d \]

4. **Setting Up the Ratio**:
We know the ratio of the 7th mean to the (m-1)th mean is \( 5:9 \):
\[ \frac{A_7}{A_{m-1}} = \frac{5}{9} \]
Substituting the expressions we found:
\[ \frac{1 + 7d}{1 + (m-1)d} = \frac{5}{9} \]

5. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 9(1 + 7d) = 5(1 + (m-1)d) \]
Expanding both sides:
\[ 9 + 63d = 5 + 5(m-1)d \]
This simplifies to:
\[ 9 + 63d = 5 + 5md - 5d \]
Rearranging gives:
\[ 9 - 5 = 5md - 5d - 63d \]
\[ 4 = 5md - 68d \]
\[ 4 = d(5m - 68) \quad \text{(Equation 2)} \]

6. **Substituting \( d \) from Equation 1**:
From Equation 1, we have:
\[ d = \frac{30}{m + 1} \]
Substituting this into Equation 2:
\[ 4 = \left(\frac{30}{m + 1}\right)(5m - 68) \]

7. **Cross-Multiplying Again**:
Cross-multiplying gives:
\[ 4(m + 1) = 30(5m - 68) \]
Expanding both sides:
\[ 4m + 4 = 150m - 2040 \]
Rearranging gives:
\[ 150m - 4m = 2044 \]
\[ 146m = 2044 \]
\[ m = \frac{2044}{146} = 14 \]

### Final Answer:
Thus, the value of \( m \) is \( 14 \).