Let three positive numbers a, b c are in geometric progression, such that a, `b+8`, c are in arithmetic progression and `a, b+8, c+64` are in geometric progression. If the arithmetic mean of a, b, c is k, then `(3)/(13)k` is equal to

Answer

To solve the problem, we need to follow the relationships given in the question step by step.

### Step 1: Establish the relationships
Given that \( a, b, c \) are in geometric progression (GP), we can write:
\[ b^2 = ac \quad \text{(1)} \]

### Step 2: Use the arithmetic progression condition
Since \( a, b+8, c \) are in arithmetic progression (AP), we can write:
\[ 2(b + 8) = a + c \quad \text{(2)} \]

### Step 3: Use the second geometric progression condition
Since \( a, b + 8, c + 64 \) are in geometric progression, we can write:
\[ (b + 8)^2 = a(c + 64) \quad \text{(3)} \]

### Step 4: Substitute \( c \) from equation (1) into equations (2) and (3)
From equation (1), we can express \( c \) as:
\[ c = \frac{b^2}{a} \quad \text{(4)} \]

Substituting (4) into equation (2):
\[ 2(b + 8) = a + \frac{b^2}{a} \]
Multiplying through by \( a \) to eliminate the fraction:
\[ 2a(b + 8) = a^2 + b^2 \]
Rearranging gives:
\[ a^2 - 2ab - 16a + b^2 = 0 \quad \text{(5)} \]

### Step 5: Substitute \( c \) into equation (3)
Substituting \( c \) from (4) into equation (3):
\[ (b + 8)^2 = a\left(\frac{b^2}{a} + 64\right) \]
This simplifies to:
\[ (b + 8)^2 = b^2 + 64a \]
Expanding the left side:
\[ b^2 + 16b + 64 = b^2 + 64a \]
Cancelling \( b^2 \) from both sides:
\[ 16b + 64 = 64a \quad \text{(6)} \]

### Step 6: Solve equations (5) and (6)
From equation (6), we can express \( a \):
\[ a = \frac{16b + 64}{64} = \frac{b + 4}{4} \quad \text{(7)} \]

### Step 7: Substitute \( a \) from (7) into (5)
Substituting (7) into (5):
\[ \left(\frac{b + 4}{4}\right)^2 - 2\left(\frac{b + 4}{4}\right)b - 16\left(\frac{b + 4}{4}\right) + b^2 = 0 \]
Multiplying through by 16 to eliminate the fractions:
\[ (b + 4)^2 - 8b(b + 4) - 64(b + 4) + 16b^2 = 0 \]
Expanding and simplifying:
\[ b^2 + 8b + 16 - 8b^2 - 32b - 64 + 16b^2 = 0 \]
Combining like terms:
\[ 9b^2 - 24b - 48 = 0 \]
Dividing by 3:
\[ 3b^2 - 8b - 16 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula:
\[ b = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-16)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ b = \frac{8 \pm \sqrt{64 + 192}}{6} = \frac{8 \pm \sqrt{256}}{6} = \frac{8 \pm 16}{6} \]
Calculating the two possible values for \( b \):
\[ b = \frac{24}{6} = 4 \quad \text{or} \quad b = \frac{-8}{6} \text{ (not valid since b is positive)} \]

### Step 9: Find \( a \) and \( c \)
Using \( b = 4 \) in equation (7):
\[ a = \frac{4 + 4}{4} = 2 \]
Using equation (1) to find \( c \):
\[ c = \frac{b^2}{a} = \frac{16}{2} = 8 \]

### Step 10: Calculate the arithmetic mean \( k \)
The arithmetic mean \( k \) of \( a, b, c \):
\[ k = \frac{a + b + c}{3} = \frac{2 + 4 + 8}{3} = \frac{14}{3} \]

### Step 11: Calculate \( \frac{3}{13}k \)
\[ \frac{3}{13}k = \frac{3}{13} \cdot \frac{14}{3} = \frac{14}{13} \]

### Final Answer
Thus, the value of \( \frac{3}{13}k \) is \( \frac{14}{13} \).

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