If a,b,c are in G.P and a,p,q are in A.P such that `2a, b+p, c +q ` are in G.P ., then the common difference of A.P is

Options

A. `sqrt(2)`a
B. `(sqrt(2)+1)(a-b)`
C. `(sqrt(2)-1)(a+b)`
D. `(sqrt(2)-1)(b-a)`

Answer

To solve the problem, we will follow a systematic approach using the properties of arithmetic and geometric progressions.

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
- We know that \( a, b, c \) are in Geometric Progression (G.P.).
- We also know that \( a, p, q \) are in Arithmetic Progression (A.P.).
- Additionally, \( 2a, b + p, c + q \) are in G.P.

2. **Using the Properties of G.P.**:
- Since \( a, b, c \) are in G.P., we can express this as:
\[ b^2 = ac \quad \text{(1)} \]

3. **Using the Properties of A.P.**:
- Since \( a, p, q \) are in A.P., we can express this as:
\[ 2p = a + q \quad \text{(2)} \]

4. **Using the Properties of the Second G.P.**:
- Since \( 2a, b + p, c + q \) are in G.P., we can express this as:
\[ (b + p)^2 = 2a(c + q) \quad \text{(3)} \]

5. **Substituting from Equation (2) into Equation (3)**:
- From equation (2), we can express \( q \) as:
\[ q = 2p - a \quad \text{(4)} \]
- Substitute \( q \) from (4) into (3):
\[ (b + p)^2 = 2a(c + (2p - a)) \]
- Simplifying this gives:
\[ (b + p)^2 = 2a(c + 2p - a) \]

6. **Substituting \( c \) from Equation (1)**:
- From equation (1), we have \( c = \frac{b^2}{a} \). Substitute this into the equation:
\[ (b + p)^2 = 2a\left(\frac{b^2}{a} + 2p - a\right) \]
- This simplifies to:
\[ (b + p)^2 = 2b^2 + 4ap - 2a^2 \]

7. **Expanding and Rearranging**:
- Expanding the left-hand side:
\[ b^2 + 2bp + p^2 = 2b^2 + 4ap - 2a^2 \]
- Rearranging gives:
\[ p^2 + 2bp - 2b^2 + 2a^2 - 4ap = 0 \]

8. **Forming a Quadratic Equation**:
- This is a quadratic equation in \( p \):
\[ p^2 + (2b - 4a)p + (2a^2 - 2b^2) = 0 \]

9. **Finding the Common Difference**:
- The common difference \( d \) of the A.P. can be found using the quadratic formula:
\[ p = \frac{-(2b - 4a) \pm \sqrt{(2b - 4a)^2 - 4(2a^2 - 2b^2)}}{2} \]
- Simplifying this will yield the common difference \( d \).

### Final Result:
The common difference \( d \) can be expressed in terms of \( a \) and \( b \) after simplifying the quadratic equation.