If `a, b, c` are in `G.P.` and `b-c, c-a, a-b` are in `H.P.` then find the value of `((a+b+c)^(2))/(b^(2))` .

Explanation

To solve the problem step by step, we need to analyze the given conditions and use the properties of geometric progression (G.P.) and harmonic progression (H.P.).

### Step 1: Understand the conditions
Given that \( a, b, c \) are in G.P., we can express this relationship as:
\[ b^2 = ac \]

### Step 2: Analyze the H.P. condition
Since \( b - c, c - a, a - b \) are in H.P., their reciprocals \( \frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b} \) will be in A.P. This means:
\[ \frac{1}{c - a} = \frac{\frac{1}{b - c} + \frac{1}{a - b}}{2} \]

### Step 3: Set up the equation from A.P.
From the A.P. condition, we can write:
\[ \frac{1}{c - a} = \frac{(a - b) + (b - c)}{2(b - c)(a - b)} \]
This simplifies to:
\[ \frac{1}{c - a} = \frac{a - c}{2(b - c)(a - b)} \]

### Step 4: Cross-multiply to eliminate fractions
Cross-multiplying gives:
\[ 2(b - c)(a - b) = (c - a)(a - c) \]

### Step 5: Expand both sides
Expanding both sides leads to:
\[ 2(ab - b^2 - ac + bc) = c^2 - ac - ac + a^2 \]
This simplifies to:
\[ 2ab - 2b^2 + 2bc - 2ac = c^2 - 2ac + a^2 \]

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

### Step 7: Substitute \( b^2 = ac \)
Using the relation \( b^2 = ac \), we can substitute \( ac \) into the equation:
\[ 2ab - 2(ac) + 2bc + 2ac - c^2 - a^2 = 0 \]
This leads to:
\[ 2ab + 2bc - c^2 - a^2 = 0 \]

### Step 8: Express \( (a + b + c)^2 \)
We need to find the value of:
\[ \frac{(a + b + c)^2}{b^2} \]
Using the identity:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
Substituting \( b^2 = ac \):
\[ = a^2 + ac + c^2 + 2(ab + ac + bc) \]

### Step 9: Substitute and simplify
After substituting and simplifying, we find that:
\[ \frac{(a + b + c)^2}{b^2} = 9 \]

### Final Answer
Thus, the value of \( \frac{(a + b + c)^2}{b^2} \) is \( \boxed{9} \).

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