Given that n arithmetic means are inserted between two sets of numbers a,2b, and 2a,b where a,b, `inR`. Suppose further that `m^(th)` mean between these two sets of numbers are same, then the ratio a:b equals

Options

A. `n-m+1:m`
B. `n-m+1:n`
C. `m:n-m+1`
D. `n:n-m+1`

Answer

To solve the problem step by step, we will analyze the given information and derive the required ratio \( \frac{a}{b} \).

### Step 1: Understanding the Problem
We have two sets of numbers:
1. The first set: \( a, A_1, A_2, \ldots, A_n, 2b \)
2. The second set: \( 2a, B_1, B_2, \ldots, B_n, b \)

We are inserting \( n \) arithmetic means between the two sets of numbers.

### Step 2: Finding the Common Difference for the First Set
The common difference \( d_1 \) for the first set can be calculated as:
\[ d_1 = \frac{(2b - a)}{n + 1} \]
The \( m^{th} \) term \( A_m \) of the first set can be expressed as:
\[ A_m = a + m \cdot d_1 = a + m \cdot \frac{(2b - a)}{n + 1} \]

### Step 3: Finding the Common Difference for the Second Set
The common difference \( d_2 \) for the second set can be calculated as:
\[ d_2 = \frac{(b - 2a)}{n + 1} \]
The \( m^{th} \) term \( B_m \) of the second set can be expressed as:
\[ B_m = 2a + m \cdot d_2 = 2a + m \cdot \frac{(b - 2a)}{n + 1} \]

### Step 4: Setting the \( m^{th} \) Terms Equal
According to the problem, the \( m^{th} \) terms of both sets are equal:
\[ A_m = B_m \]
Substituting the expressions we found:
\[ a + m \cdot \frac{(2b - a)}{n + 1} = 2a + m \cdot \frac{(b - 2a)}{n + 1} \]

### Step 5: Rearranging the Equation
Rearranging the equation gives us:
\[ a + m \cdot \frac{(2b - a)}{n + 1} - 2a - m \cdot \frac{(b - 2a)}{n + 1} = 0 \]
This simplifies to:
\[ -a + m \left( \frac{(2b - a) - (b - 2a)}{n + 1} \right) = 0 \]
\[ -a + m \left( \frac{(2b - a - b + 2a)}{n + 1} \right) = 0 \]
\[ -a + m \left( \frac{(b + a)}{n + 1} \right) = 0 \]

### Step 6: Solving for the Ratio \( \frac{a}{b} \)
From the equation:
\[ -a = -m \cdot \frac{(b + a)}{n + 1} \]
We can rearrange this to find:
\[ a(n + 1) = m(b + a) \]
This leads to:
\[ a(n + 1) - mb = ma \]
Rearranging gives:
\[ a(n + 1 - m) = mb \]
Thus, we find the ratio:
\[ \frac{a}{b} = \frac{m}{n + 1 - m} \]

### Final Answer
The ratio \( \frac{a}{b} \) is:
\[ \frac{a}{b} = \frac{m}{n + 1 - m} \]