n A.M.\'s are inserted between 1 and 31 such that the ratio of the 7th and `(n-1)` th means is 5:9. Find n.

Options

A. 12
B. 13
C. 14
D. 15

Answer

To solve the problem of finding the value of \( n \) when \( n \) Arithmetic Means (A.M.s) are inserted between 1 and 31 such that the ratio of the 7th and \( (n-1) \)th means is \( 5:9 \), we can follow these steps:

### Step 1: Understand the Problem
We need to insert \( n \) A.M.s between the numbers 1 and 31. This means we will have a total of \( n + 2 \) terms (1, the \( n \) A.M.s, and 31).

### Step 2: Identify the First and Last Terms
Let \( a = 1 \) (the first term) and \( l = 31 \) (the last term). The total number of terms is \( n + 2 \).

### Step 3: Calculate the Common Difference \( d \)
The formula for the \( m \)th term of an A.P. (Arithmetic Progression) is given by:
\[ T_m = a + (m-1)d \]
For our case, the last term can be expressed as:
\[ T_{n+2} = a + (n+1)d = 31 \]
Substituting \( a = 1 \):
\[ 1 + (n+1)d = 31 \]
This simplifies to:
\[ (n+1)d = 30 \quad \text{(Equation 1)} \]

### Step 4: Express the Terms
The 7th term \( T_7 \) and the \( (n-1) \)th term \( T_{n-1} \) can be expressed as:
\[ T_7 = a + 6d = 1 + 6d \]
\[ T_{n-1} = a + (n-2)d = 1 + (n-2)d \]

### Step 5: Set Up the Ratio
According to the problem, the ratio of the 7th term to the \( (n-1) \)th term is:
\[ \frac{T_7}{T_{n-1}} = \frac{5}{9} \]
Substituting the expressions for \( T_7 \) and \( T_{n-1} \):
\[ \frac{1 + 6d}{1 + (n-2)d} = \frac{5}{9} \]

### Step 6: Cross-Multiply
Cross-multiplying gives:
\[ 9(1 + 6d) = 5(1 + (n-2)d) \]
Expanding both sides:
\[ 9 + 54d = 5 + 5(n-2)d \]
This simplifies to:
\[ 9 + 54d = 5 + 5nd - 10d \]
Combining like terms:
\[ 9 + 54d = 5 + 5nd - 10d \]
\[ 9 + 64d = 5 + 5nd \]
Rearranging gives:
\[ 64d - 5nd = -4 \]
Factoring out \( d \):
\[ d(64 - 5n) = -4 \quad \text{(Equation 2)} \]

### Step 7: Substitute \( d \) from Equation 1
From Equation 1, we have:
\[ d = \frac{30}{n+1} \]
Substituting this into Equation 2:
\[ \frac{30}{n+1}(64 - 5n) = -4 \]
Cross-multiplying gives:
\[ 30(64 - 5n) = -4(n + 1) \]
Expanding both sides:
\[ 1920 - 150n = -4n - 4 \]
Combining like terms:
\[ 1920 + 4 = 150n - 4n \]
\[ 1924 = 146n \]
Dividing both sides by 146:
\[ n = \frac{1924}{146} = 13.18 \quad \text{(approximately)} \]

### Step 8: Conclusion
Since \( n \) must be a whole number, we round \( n \) to the nearest whole number, which is 14.

Thus, the value of \( n \) is **14**.