If `a^2+b^2,a b+b c ,a n db^2+c^2` are in G.P., then `a ,b ,c` are in a. A.P. b. G.P. c. H.P. d. none of these

Options

A. A.P.
B. G.P
C. H.P
D. none of these

Explanation

To solve the problem, we need to determine the relationship between \(a\), \(b\), and \(c\) given that \(a^2 + b^2\), \(ab + bc\), and \(b^2 + c^2\) are in geometric progression (G.P.).

### Step 1: Understanding the condition for G.P.
For three terms \(x\), \(y\), and \(z\) to be in G.P., the condition is:
\[ y^2 = x \cdot z \]
In our case, let:
- \(x = a^2 + b^2\)
- \(y = ab + bc\)
- \(z = b^2 + c^2\)

### Step 2: Setting up the equation
According to the G.P. condition, we have:
\[ (ab + bc)^2 = (a^2 + b^2)(b^2 + c^2) \]

### Step 3: Expanding both sides
Now, we will expand both sides of the equation.

**Left Side:**
\[ (ab + bc)^2 = a^2b^2 + 2ab \cdot bc + b^2c^2 \]

**Right Side:**
\[ (a^2 + b^2)(b^2 + c^2) = a^2b^2 + a^2c^2 + b^4 + b^2c^2 \]

### Step 4: Equating the two sides
Now we equate the two expansions:
\[ a^2b^2 + 2ab \cdot bc + b^2c^2 = a^2b^2 + a^2c^2 + b^4 + b^2c^2 \]

### Step 5: Simplifying the equation
We can cancel \(a^2b^2\) and \(b^2c^2\) from both sides:
\[ 2ab \cdot bc = a^2c^2 + b^4 \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ b^4 - 2abc + a^2c^2 = 0 \]

### Step 7: Recognizing the quadratic form
This is a quadratic equation in terms of \(b^2\):
\[ b^4 - 2abc + a^2c^2 = 0 \]
Let \(x = b^2\), then we have:
\[ x^2 - 2acx + a^2c^2 = 0 \]

### Step 8: Finding the roots
The discriminant of this quadratic must be non-negative for \(b^2\) to have real solutions:
\[ D = (2ac)^2 - 4 \cdot 1 \cdot a^2c^2 = 4a^2c^2 - 4a^2c^2 = 0 \]
This means the quadratic has a double root, indicating:
\[ b^2 = ac \]

### Step 9: Conclusion about \(a\), \(b\), and \(c\)
Since \(b^2 = ac\), we can conclude that \(a\), \(b\), and \(c\) are in geometric progression (G.P.).

### Final Answer
The answer is **b. G.P.**

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