In a triangle ABC, 2B= A + C and `b^(2) = ac`, then `(a(a+b+c))/(3bc)` is

Answer

To solve the problem step by step, we start with the given conditions in triangle ABC.

### Step 1: Understand the given conditions
We have two conditions:
1. \( 2B = A + C \)
2. \( b^2 = ac \)

From the first condition, we can express \( A + C \) in terms of \( B \):
\[ A + C = 2B \]

### Step 2: Use the angle sum property of triangles
The sum of angles in a triangle is 180 degrees:
\[ A + B + C = 180^\circ \]
Substituting \( A + C = 2B \) into this equation gives:
\[ 2B + B = 180^\circ \implies 3B = 180^\circ \implies B = 60^\circ \]

### Step 3: Find angles A and C
Since \( B = 60^\circ \), we can substitute back to find \( A \) and \( C \):
\[ A + C = 2B = 120^\circ \]
Thus, we can express \( C \) as:
\[ C = 120^\circ - A \]

### Step 4: Use the second condition \( b^2 = ac \)
Using the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Let \( k \) be the common ratio:
\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]
Substituting \( B = 60^\circ \):
\[ b = k \sin 60^\circ = k \frac{\sqrt{3}}{2} \]
Now substituting into the second condition:
\[ b^2 = ac \implies \left(k \frac{\sqrt{3}}{2}\right)^2 = (k \sin A)(k \sin C) \]
This simplifies to:
\[ \frac{3k^2}{4} = k^2 \sin A \sin C \]
Dividing both sides by \( k^2 \) (assuming \( k \neq 0 \)):
\[ \frac{3}{4} = \sin A \sin C \]

### Step 5: Substitute \( C = 120^\circ - A \)
Using the identity \( \sin(120^\circ - A) = \sin 120^\circ \cos A - \cos 120^\circ \sin A \):
\[ \sin C = \sin(120^\circ - A) = \frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A \]
Thus, we have:
\[ \sin A \left( \frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A \right) = \frac{3}{4} \]

### Step 6: Solve for \( A \) and \( C \)
Using the sine values and solving the equation leads to the conclusion that \( A = C = 60^\circ \). Therefore, triangle ABC is equilateral:
\[ A = B = C = 60^\circ \]

### Step 7: Calculate the expression \( \frac{a(a+b+c)}{3bc} \)
In an equilateral triangle, \( a = b = c \). Let \( a = b = c = k \):
\[ \frac{a(a+b+c)}{3bc} = \frac{k(k+k+k)}{3kk} = \frac{k(3k)}{3k^2} = \frac{3k^2}{3k^2} = 1 \]

### Final Answer
Thus, the value of \( \frac{a(a+b+c)}{3bc} \) is:
\[ \boxed{1} \]