If a, b and c are in G.P and x and y, respectively , be arithmetic means between a,b and b,c then

Options

A. `a/x+c/y=2`
B. `a/x+c/y=c/a`
C. `1/x +1/y=2/b`
D. `1/x+1/y=2/ac`

Explanation

To solve the problem, we need to verify the two expressions involving \( a, b, c \) (which are in geometric progression) and \( x, y \) (which are arithmetic means).

### Step-by-step Solution:

1. **Understanding the Given Information:**
- Since \( a, b, c \) are in geometric progression (G.P.), we have:
\[ b^2 = ac \]

2. **Finding the Arithmetic Means:**
- The arithmetic mean \( x \) between \( a \) and \( b \) is given by:
\[ x = \frac{a + b}{2} \]
- The arithmetic mean \( y \) between \( b \) and \( c \) is given by:
\[ y = \frac{b + c}{2} \]

3. **Finding the Expression \( \frac{a}{x} + \frac{c}{y} \):**
- Substitute the values of \( x \) and \( y \):
\[ \frac{a}{x} = \frac{a}{\frac{a + b}{2}} = \frac{2a}{a + b} \]
\[ \frac{c}{y} = \frac{c}{\frac{b + c}{2}} = \frac{2c}{b + c} \]
- Now, adding these two fractions:
\[ \frac{a}{x} + \frac{c}{y} = \frac{2a}{a + b} + \frac{2c}{b + c} \]
- To add these fractions, we need a common denominator:
\[ = \frac{2a(b + c) + 2c(a + b)}{(a + b)(b + c)} \]
- Simplifying the numerator:
\[ = \frac{2ab + 2ac + 2ca + 2cb}{(a + b)(b + c)} = \frac{2(ab + ac + bc)}{(a + b)(b + c)} \]

4. **Verifying the Expression \( \frac{1}{x} + \frac{1}{y} \):**
- We can find \( \frac{1}{x} \) and \( \frac{1}{y} \):
\[ \frac{1}{x} = \frac{2}{a + b} \]
\[ \frac{1}{y} = \frac{2}{b + c} \]
- Adding these gives:
\[ \frac{1}{x} + \frac{1}{y} = \frac{2}{a + b} + \frac{2}{b + c} \]
- Again, finding a common denominator:
\[ = \frac{2(b + c) + 2(a + b)}{(a + b)(b + c)} = \frac{2(b + c + a + b)}{(a + b)(b + c)} = \frac{2(a + 2b + c)}{(a + b)(b + c)} \]

### Final Results:
- We have shown that:
- \( \frac{a}{x} + \frac{c}{y} = \frac{2(ab + ac + bc)}{(a + b)(b + c)} \)
- \( \frac{1}{x} + \frac{1}{y} = \frac{2(a + 2b + c)}{(a + b)(b + c)} \)