`A_(1), A_(2)` are two arithmetic means between two positive real numbers a and b. P is the point of intersection of the perpendicular lines `2x+y-20=0` and `x-2y+10=0`. If the centroid of the triangle with vertices `(a, b), (A_(1), A_(2))` and `(A_(2), A_(1))` lies at the point P, the value of ab is equal to

Answer

To solve the problem step by step, we will follow the outlined logic and calculations based on the information provided.

### Step 1: Identify the Arithmetic Means
Let \( A_1 \) and \( A_2 \) be the two arithmetic means between two positive real numbers \( a \) and \( b \). The arithmetic means can be expressed as:
- \( A_1 = a + d \)
- \( A_2 = a + 2d \)
- \( b = a + 3d \)

Where \( d \) is the common difference.

### Step 2: Find the Intersection Point P
We need to find the point of intersection of the lines given by the equations:
1. \( 2x + y - 20 = 0 \)
2. \( x - 2y + 10 = 0 \)

To find the intersection, we can solve these equations simultaneously.

From the first equation:
\[ y = 20 - 2x \]

Substituting \( y \) in the second equation:
\[ x - 2(20 - 2x) + 10 = 0 \]
\[ x - 40 + 4x + 10 = 0 \]
\[ 5x - 30 = 0 \]
\[ x = 6 \]

Now substituting \( x = 6 \) back into the first equation to find \( y \):
\[ y = 20 - 2(6) = 20 - 12 = 8 \]

Thus, the point \( P \) is \( (6, 8) \).

### Step 3: Find the Centroid of the Triangle
The vertices of the triangle are \( (a, b) \), \( (A_1, A_2) \), and \( (A_2, A_1) \). The coordinates of the centroid \( G \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \) is given by:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

Substituting the vertices:
- \( x_1 = a \)
- \( x_2 = A_1 = a + d \)
- \( x_3 = A_2 = a + 2d \)

The x-coordinate of the centroid:
\[ \frac{a + (a + d) + (a + 2d)}{3} = \frac{3a + 3d}{3} = a + d \]

For the y-coordinates:
- \( y_1 = b \)
- \( y_2 = A_1 = a + d \)
- \( y_3 = A_2 = a + 2d \)

The y-coordinate of the centroid:
\[ \frac{b + (a + d) + (a + 2d)}{3} = \frac{b + 2a + 3d}{3} \]

### Step 4: Set the Centroid Equal to Point P
Since the centroid lies at point \( P(6, 8) \), we can set up the following equations:
1. \( a + d = 6 \)
2. \( \frac{b + 2a + 3d}{3} = 8 \)

From the first equation:
\[ d = 6 - a \]

Substituting \( d \) into the second equation:
\[ b + 2a + 3(6 - a) = 24 \]
\[ b + 2a + 18 - 3a = 24 \]
\[ b - a + 18 = 24 \]
\[ b - a = 6 \]

### Step 5: Substitute for b
From the relationship \( b = a + 6 \) and substituting \( b \) back into the equation for \( b \):
\[ b = a + 3d \]
Substituting \( d = 6 - a \):
\[ b = a + 3(6 - a) = a + 18 - 3a = 18 - 2a \]

Setting the two expressions for \( b \) equal:
\[ a + 6 = 18 - 2a \]
\[ 3a = 12 \]
\[ a = 4 \]

### Step 6: Find b
Now substituting \( a \) back to find \( b \):
\[ b = 4 + 6 = 10 \]

### Step 7: Calculate ab
Finally, we calculate \( ab \):
\[ ab = 4 \times 10 = 40 \]

Thus, the value of \( ab \) is \( \boxed{40} \).

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