Ifa,b,c are in A.P., a,x,b,are in G.P and b,y,c are also in G.P then the point (x,y) lies on

Options

A. a line
B. a circle
C. an ellipse
D. a hyperbola

Answer

To solve the problem step by step, we need to analyze the given conditions about the sequences and derive the relationship between \(x\) and \(y\).

### Step 1: Understanding the Given Conditions
We know that:
1. \(a, b, c\) are in Arithmetic Progression (A.P.).
2. \(a, x, b\) are in Geometric Progression (G.P.).
3. \(b, y, c\) are also in Geometric Progression (G.P.).

### Step 2: Using the A.P. Condition
Since \(a, b, c\) are in A.P., we can express this condition mathematically:
\[ b = \frac{a + c}{2} \]
This is our **Equation (1)**.

### Step 3: Using the G.P. Condition for \(a, x, b\)
For \(a, x, b\) to be in G.P., the relationship can be expressed as:
\[ x^2 = ab \]
This implies:
\[ x = \sqrt{ab} \]
This is our **Equation (2)**.

### Step 4: Using the G.P. Condition for \(b, y, c\)
For \(b, y, c\) to be in G.P., we have:
\[ y^2 = bc \]
This implies:
\[ y = \sqrt{bc} \]
This is our **Equation (3)**.

### Step 5: Expressing \(c\) in terms of \(a\) and \(b\)
From Equation (1), we can express \(c\):
\[ c = 2b - a \]

### Step 6: Substituting \(c\) in Equation (3)
Now, substituting \(c\) into Equation (3):
\[ y^2 = b(2b - a) \]
This simplifies to:
\[ y^2 = 2b^2 - ab \]

### Step 7: Substituting \(c\) in Equation (2)
Now, substituting \(c\) into Equation (2):
\[ x^2 = ab \]

### Step 8: Combining the Equations
Now we have:
1. \(x^2 = ab\) (from Equation (2))
2. \(y^2 = 2b^2 - ab\) (from Equation (3))

### Step 9: Expressing \(y^2\) in terms of \(x^2\)
From \(x^2 = ab\), we can substitute \(ab\) in the equation for \(y^2\):
\[ y^2 = 2b^2 - x^2 \]

### Step 10: Rearranging the Equation
Rearranging gives us:
\[ x^2 + y^2 = 2b^2 \]

### Step 11: Recognizing the Equation of a Circle
The equation \(x^2 + y^2 = 2b^2\) represents a circle centered at the origin \((0, 0)\) with a radius of \(\sqrt{2}b\).

### Conclusion
Thus, the point \((x, y)\) lies on the circle given by the equation:
\[ x^2 + y^2 = 2b^2 \]