If `5a-b, 2a+b, a+2b` are in A.P. and `(a-1)^(2), (ab+1),(b+1)^(2)` are in G.P., `a ne 0`, then a is equal to

Options

A. `2,-1//4`
B. `-2,-1//4`
C. `-2,1//4`
D. `2,1//4`

Answer

To solve the problem step by step, we need to analyze the conditions given in the question.

### Step 1: Understanding the conditions
We have two sets of numbers:
1. \(5a - b\), \(2a + b\), \(a + 2b\) are in Arithmetic Progression (A.P.)
2. \((a - 1)^2\), \(ab + 1\), \((b + 1)^2\) are in Geometric Progression (G.P.)

### Step 2: Using the A.P. condition
For three numbers \(A\), \(B\), \(C\) to be in A.P., the condition is:
\[ 2B = A + C \]

Here, let:
- \(A = 5a - b\)
- \(B = 2a + b\)
- \(C = a + 2b\)

Applying the A.P. condition:
\[ 2(2a + b) = (5a - b) + (a + 2b) \]

Expanding both sides:
\[ 4a + 2b = 5a - b + a + 2b \]
\[ 4a + 2b = 6a + b \]

### Step 3: Rearranging the equation
Now, rearranging the equation:
\[ 4a + 2b - 6a - b = 0 \]
\[ -2a + b = 0 \quad \Rightarrow \quad b = 2a \]

### Step 4: Using the G.P. condition
Now, we use the G.P. condition. For three numbers \(X\), \(Y\), \(Z\) to be in G.P., the condition is:
\[ Y^2 = XZ \]

Let:
- \(X = (a - 1)^2\)
- \(Y = ab + 1\)
- \(Z = (b + 1)^2\)

Applying the G.P. condition:
\[ (ab + 1)^2 = (a - 1)^2 \cdot (b + 1)^2 \]

### Step 5: Substituting \(b\) in G.P. condition
Substituting \(b = 2a\) into the G.P. condition:
\[ (2a^2 + 1)^2 = (a - 1)^2 \cdot (2a + 1)^2 \]

### Step 6: Expanding both sides
Expanding the left-hand side:
\[ (2a^2 + 1)^2 = 4a^4 + 4a^2 + 1 \]

Expanding the right-hand side:
\[ (a - 1)^2 = a^2 - 2a + 1 \]
\[ (2a + 1)^2 = 4a^2 + 4a + 1 \]
Now, multiplying:
\[ (a^2 - 2a + 1)(4a^2 + 4a + 1) = 4a^4 + 4a^3 + a^2 - 8a^3 - 8a^2 - 2a + 4a^2 + 4a + 1 \]
Combining like terms:
\[ 4a^4 - 4a^3 - 3a^2 + 2a + 1 \]

### Step 7: Setting the equation
Now we set the two sides equal:
\[ 4a^4 + 4a^2 + 1 = 4a^4 - 4a^3 - 3a^2 + 2a + 1 \]
Cancelling \(4a^4\) and \(1\) from both sides:
\[ 4a^2 = -4a^3 - 3a^2 + 2a \]
Rearranging gives:
\[ 4a^3 + 7a^2 - 2a = 0 \]

### Step 8: Factoring out \(a\)
Factoring out \(a\):
\[ a(4a^2 + 7a - 2) = 0 \]
Since \(a \neq 0\), we solve:
\[ 4a^2 + 7a - 2 = 0 \]

### Step 9: Using the quadratic formula
Using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 + 32}}{8} \]
\[ = \frac{-7 \pm 9}{8} \]
Calculating the two possible values:
1. \(a = \frac{2}{8} = \frac{1}{4}\)
2. \(a = \frac{-16}{8} = -2\)

### Conclusion
Since \(a \neq 0\), the possible values of \(a\) are:
- \(a = \frac{1}{4}\)
- \(a = -2\)

Thus, the final answer is:
\[ \text{The value of } a \text{ is } \frac{1}{4} \text{ or } -2. \]