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 find the value of \( \frac{3}{13}k \) where \( k \) is the arithmetic mean of three positive numbers \( a, b, c \) that meet specific conditions. Let's break down the solution step by step.

### Step 1: Understanding the Conditions
We know that:
1. \( a, b, c \) are in geometric progression (GP).
2. \( a, b + 8, c \) are in arithmetic progression (AP).
3. \( a, b + 8, c + 64 \) are in geometric progression (GP).

### Step 2: Expressing the GP Condition
Since \( a, b, c \) are in GP, we can express this condition as:
\[ b^2 = ac \quad \text{(1)} \]

### Step 3: Expressing the AP Condition
For the numbers \( a, b + 8, c \) to be in AP, we have:
\[ 2(b + 8) = a + c \quad \text{(2)} \]

### Step 4: Expressing the Second GP Condition
For the numbers \( a, b + 8, c + 64 \) to be in GP, we have:
\[ (b + 8)^2 = a(c + 64) \quad \text{(3)} \]

### Step 5: Substituting and Simplifying
From equation (1), we know \( ac = b^2 \). We can substitute \( ac \) in equation (3):
\[ (b + 8)^2 = b^2 + 64a \quad \text{(substituting \( ac \) from (1))} \]
Expanding the left side:
\[ b^2 + 16b + 64 = b^2 + 64a \]
Cancelling \( b^2 \) from both sides:
\[ 16b + 64 = 64a \quad \text{(4)} \]

### Step 6: Using Equation (2) and (4)
From equation (2):
\[ 2b + 16 = a + c \quad \text{(5)} \]
Now, we can express \( c \) in terms of \( a \) and \( b \):
\[ c = 2b + 16 - a \quad \text{(6)} \]

### Step 7: Substitute \( c \) in Equation (1)
Substituting \( c \) from (6) into (1):
\[ b^2 = a(2b + 16 - a) \]
Expanding this:
\[ b^2 = 2ab + 16a - a^2 \]
Rearranging gives:
\[ a^2 - (2b + 16)a + b^2 = 0 \quad \text{(7)} \]

### Step 8: Solving for \( a \)
Using the quadratic formula on equation (7):
\[ a = \frac{(2b + 16) \pm \sqrt{(2b + 16)^2 - 4b^2}}{2} \]
Calculating the discriminant:
\[ (2b + 16)^2 - 4b^2 = 4b^2 + 64b + 256 - 4b^2 = 64b + 256 \]
Thus:
\[ a = \frac{(2b + 16) \pm \sqrt{64(b + 4)}}{2} \]
This simplifies to:
\[ a = b + 8 \pm 4\sqrt{b + 4} \]

### Step 9: Finding \( b \)
Using equation (4):
\[ 16b + 64 = 64a \]
Substituting \( a \):
\[ 16b + 64 = 64(b + 8 \pm 4\sqrt{b + 4}) \]
This can be solved for \( b \).

### Step 10: Finding \( k \)
Once we find \( a, b, c \), we can calculate the arithmetic mean \( k \):
\[ k = \frac{a + b + c}{3} \]

### Step 11: Calculate \( \frac{3}{13}k \)
Finally, substitute the value of \( k \) into \( \frac{3}{13}k \) to find the answer.