Find three consecutive numbers in G.P. whose sum is 28 and product is 512.

Explanation

To find three consecutive numbers in Geometric Progression (G.P.) whose sum is 28 and product is 512, we can follow these steps:

### Step 1: Define the terms in G.P.
Let the three consecutive terms in G.P. be:
- First term: \( \frac{a}{r} \)
- Second term: \( a \)
- Third term: \( ar \)

### Step 2: Set up the equations based on the given conditions
According to the problem, we have two conditions:
1. The sum of the terms is 28:
\[ \frac{a}{r} + a + ar = 28 \]
2. The product of the terms is 512:
\[ \left(\frac{a}{r}\right) \cdot a \cdot (ar) = 512 \]

### Step 3: Simplify the product equation
The product simplifies as follows:
\[ \frac{a^3}{r} = 512 \]
From this, we can express \( a^3 \):
\[ a^3 = 512r \]

### Step 4: Substitute \( a^3 \) in the sum equation
From the sum equation, we can factor out \( a \):
\[ a \left(\frac{1}{r} + 1 + r\right) = 28 \]
Now, substituting \( a = \sqrt[3]{512r} \) into the sum equation:
\[ \sqrt[3]{512r} \left(\frac{1}{r} + 1 + r\right) = 28 \]

### Step 5: Solve for \( a \) and \( r \)
To simplify, we can first find \( a \) from the product equation:
Taking the cube root of both sides:
\[ a = \sqrt[3]{512r} = 8\sqrt[3]{r} \]

### Step 6: Substitute \( a \) back into the sum equation
Substituting \( a \) back into the sum equation:
\[ 8\sqrt[3]{r} \left(\frac{1}{r} + 1 + r\right) = 28 \]
Dividing both sides by 8:
\[ \sqrt[3]{r} \left(\frac{1}{r} + 1 + r\right) = 3.5 \]

### Step 7: Clear the fraction
Multiply through by \( r \):
\[ \sqrt[3]{r} (1 + r + r^2) = 3.5r \]

### Step 8: Rearranging and forming a quadratic equation
This leads to a cubic equation in \( r \). To find \( r \), we can solve the quadratic formed from the rearranged sum equation:
\[ 2r^2 - 5r + 2 = 0 \]

### Step 9: Factor the quadratic equation
Factoring gives:
\[ (2r - 1)(r - 2) = 0 \]
Thus, we have two possible values for \( r \):
1. \( r = \frac{1}{2} \)
2. \( r = 2 \)

### Step 10: Find corresponding \( a \) values
For \( r = 2 \):
\[ a = 8 \]
The terms are:
- \( \frac{8}{2} = 4 \)
- \( 8 \)
- \( 8 \cdot 2 = 16 \)

For \( r = \frac{1}{2} \):
\[ a = 8 \]
The terms are:
- \( \frac{8}{\frac{1}{2}} = 16 \)
- \( 8 \)
- \( 8 \cdot \frac{1}{2} = 4 \)

### Final Answer
The three consecutive numbers in G.P. are:
- \( 4, 8, 16 \) or \( 16, 8, 4 \)