Let `a ,b >0,` let `5a-b ,2a+b ,a+2b` be in A.P. and `(b+1)^2, a b+1,(a-1)^2` are in G.P., then the value of `(a^(-1)+b^(-1))` is _______.

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Set up the equations for A.P.
Given that \(5a - b\), \(2a + b\), and \(a + 2b\) are in Arithmetic Progression (A.P.), we can use the property of A.P. which states that the middle term is the average of the other two terms. Therefore, we have:

\[ 2(2a + b) = (5a - b) + (a + 2b) \]

### Step 2: Simplify the A.P. equation
Expanding the equation from Step 1:

\[ 4a + 2b = 5a - b + a + 2b \]

Combining like terms:

\[ 4a + 2b = 6a + b \]

### Step 3: Rearranging the equation
Rearranging the equation gives:

\[ 4a + 2b - 6a - b = 0 \]

This simplifies to:

\[ -2a + b = 0 \quad \Rightarrow \quad b = 2a \]

### Step 4: Set up the equations for G.P.
Next, we know that \((b + 1)^2\), \(ab + 1\), and \((a - 1)^2\) are in Geometric Progression (G.P.). The property of G.P. states that the square of the middle term is equal to the product of the other two terms:

\[ (ab + 1)^2 = (b + 1)(a - 1)^2 \]

### Step 5: Substitute \(b\) into the G.P. equation
Substituting \(b = 2a\) into the G.P. equation gives:

\[ (2a + 1)^2 = (2a + 1)(a - 1)^2 \]

### Step 6: Expand both sides
Expanding both sides:

Left-hand side:

\[ (2a + 1)^2 = 4a^2 + 4a + 1 \]

Right-hand side:

\[ (2a + 1)(a^2 - 2a + 1) = 2a^3 - 4a^2 + 2a + a^2 - 2a + 1 = 2a^3 - 3a^2 + 1 \]

### Step 7: Set the equations equal to each other
Setting the left-hand side equal to the right-hand side:

\[ 4a^2 + 4a + 1 = 2a^3 - 3a^2 + 1 \]

### Step 8: Rearranging the equation
Rearranging gives:

\[ 2a^3 - 7a^2 - 4a = 0 \]

### Step 9: Factor the equation
Factoring out \(a\):

\[ a(2a^2 - 7a - 4) = 0 \]

### Step 10: Solve the quadratic equation
We can solve the quadratic equation \(2a^2 - 7a - 4 = 0\) using the quadratic formula:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} \]

Calculating the discriminant:

\[ 49 + 32 = 81 \quad \Rightarrow \quad \sqrt{81} = 9 \]

Thus:

\[ a = \frac{7 \pm 9}{4} \]

This gives us two potential solutions for \(a\):

1. \(a = \frac{16}{4} = 4\)
2. \(a = \frac{-2}{4} = -\frac{1}{2}\) (not valid since \(a > 0\))

So, \(a = 4\).

### Step 11: Find \(b\)
Using \(b = 2a\):

\[ b = 2 \cdot 4 = 8 \]

### Step 12: Calculate \(\frac{1}{a} + \frac{1}{b}\)
Now, we need to find \(\frac{1}{a} + \frac{1}{b}\):

\[ \frac{1}{a} + \frac{1}{b} = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8} \]

### Final Answer
Thus, the value of \(\frac{1}{a} + \frac{1}{b}\) is \(\frac{3}{8}\).

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