If a, b, c are in G.P., x and y are the A.M.'s of a, b and b, c respectively, then prove that:
`(i)(a)/(x)+(c)/(y)=2" "(ii) (1)/(x)+(1)/(y)=(2)/(b)`

Explanation

To solve the given problem step by step, let's break it down into two parts as required.

### Given:
- \( a, b, c \) are in Geometric Progression (G.P.).
- \( x \) is the Arithmetic Mean (A.M.) of \( a \) and \( b \).
- \( y \) is the A.M. of \( b \) and \( c \).

### To Prove:
1. \( \frac{a}{x} + \frac{c}{y} = 2 \)
2. \( \frac{1}{x} + \frac{1}{y} = \frac{2}{b} \)

### Step 1: Understanding the G.P. condition
Since \( a, b, c \) are in G.P., we have:
\[ b^2 = ac \tag{1} \]

### Step 2: Finding the Arithmetic Means
The arithmetic mean \( x \) of \( a \) and \( b \) is given by:
\[ x = \frac{a + b}{2} \tag{2} \]
The arithmetic mean \( y \) of \( b \) and \( c \) is given by:
\[ y = \frac{b + c}{2} \tag{3} \]

### Step 3: Proving \( \frac{a}{x} + \frac{c}{y} = 2 \)
We start with the left-hand side (LHS):
\[ \text{LHS} = \frac{a}{x} + \frac{c}{y} \]
Substituting \( x \) and \( y \) from equations (2) and (3):
\[ = \frac{a}{\frac{a + b}{2}} + \frac{c}{\frac{b + c}{2}} \]
This simplifies to:
\[ = \frac{2a}{a + b} + \frac{2c}{b + c} \]
Taking a common denominator:
\[ = \frac{2a(b + c) + 2c(a + b)}{(a + b)(b + c)} \]
Expanding the numerator:
\[ = \frac{2ab + 2ac + 2ca + 2bc}{(a + b)(b + c)} \]
This simplifies to:
\[ = \frac{2(ab + ac + bc)}{(a + b)(b + c)} \]

Now, using equation (1) where \( b^2 = ac \):
\[ = \frac{2(ab + b^2 + bc)}{(a + b)(b + c)} \]
Substituting \( ac \) with \( b^2 \):
\[ = \frac{2(ab + b^2 + bc)}{(a + b)(b + c)} = 2 \]

### Step 4: Proving \( \frac{1}{x} + \frac{1}{y} = \frac{2}{b} \)
Now, we start with the left-hand side (LHS):
\[ \text{LHS} = \frac{1}{x} + \frac{1}{y} \]
Substituting \( x \) and \( y \):
\[ = \frac{2}{a + b} + \frac{2}{b + c} \]
Taking a common denominator:
\[ = \frac{2(b + c) + 2(a + b)}{(a + b)(b + c)} \]
This simplifies to:
\[ = \frac{2b + 2c + 2a + 2b}{(a + b)(b + c)} = \frac{2(a + 2b + c)}{(a + b)(b + c)} \]

Now, using \( b^2 = ac \):
\[ = \frac{2(a + 2b + c)}{(a + b)(b + c)} = \frac{2}{b} \]

### Conclusion
Thus, we have proved both parts:
1. \( \frac{a}{x} + \frac{c}{y} = 2 \)
2. \( \frac{1}{x} + \frac{1}{y} = \frac{2}{b} \)