n arithmetic means are inserted in between x and 2y and then between 2x and y In case the rth mean in each case be equal , then find the ration `x/y`

Answer

To solve the problem, we need to find the ratio \( \frac{x}{y} \) given that \( n \) arithmetic means are inserted between \( x \) and \( 2y \), and between \( 2x \) and \( y \), and that the \( r \)-th means in both cases are equal.

### Step-by-Step Solution:

1. **Identify the first series**: We have \( x \) and \( 2y \) with \( n \) arithmetic means inserted between them. The total number of terms in this series is \( n + 2 \) (including \( x \) and \( 2y \)).

2. **Find the common difference \( D_1 \)**: The common difference \( D_1 \) can be calculated using the formula for the last term of an arithmetic series:
\[ 2y = x + (n + 1)D_1 \]
Rearranging gives:
\[ D_1 = \frac{2y - x}{n + 1} \]

3. **Identify the second series**: Now consider the series with \( 2x \) and \( y \), again with \( n \) arithmetic means inserted. The total number of terms is also \( n + 2 \).

4. **Find the common difference \( D_2 \)**: Similarly, we can find \( D_2 \) using:
\[ y = 2x + (n + 1)D_2 \]
Rearranging gives:
\[ D_2 = \frac{y - 2x}{n + 1} \]

5. **Find the \( r \)-th mean in both series**: The \( r \)-th mean in the first series is given by:
\[ A_r = x + rD_1 = x + r\left(\frac{2y - x}{n + 1}\right) \]
The \( r \)-th mean in the second series is:
\[ B_r = 2x + rD_2 = 2x + r\left(\frac{y - 2x}{n + 1}\right) \]

6. **Set the two means equal**: Since the \( r \)-th means are equal, we have:
\[ x + r\left(\frac{2y - x}{n + 1}\right) = 2x + r\left(\frac{y - 2x}{n + 1}\right) \]

7. **Clear the fractions**: Multiply through by \( n + 1 \) to eliminate the denominator:
\[ (n + 1)x + r(2y - x) = (n + 1)(2x) + r(y - 2x) \]

8. **Expand and simplify**:
\[ (n + 1)x + 2ry - rx = 2(n + 1)x + ry - 2rx \]
Rearranging gives:
\[ (n + 1)x + 2ry - rx - 2(n + 1)x - ry + 2rx = 0 \]
Combine like terms:
\[ -nx + ry + (r - 2)(x) = 0 \]

9. **Rearranging**: This can be rearranged to express \( x \) in terms of \( y \):
\[ (n + 1 - r)x = ry \]

10. **Find the ratio \( \frac{x}{y} \)**:
\[ \frac{x}{y} = \frac{r}{n + 1 - r} \]

### Final Answer:
Thus, the ratio \( \frac{x}{y} \) is given by:
\[ \frac{x}{y} = \frac{r}{n + 1 - r} \]