Three numbers are in G.P. Their sum is 28 and product is 512. The numbers are

Options

A. (a) 4, 8, 16
B. (b) 6, 12, 24
C. (c) 3, 9, 27
D. (d) 2, 4, 16

Answer

To find the three numbers in a geometric progression (G.P.) whose sum is 28 and product is 512, we can follow these steps:

### Step 1: Define the terms of the G.P.
Let the three numbers in G.P. be:
- First term: \( a/r \)
- Second term: \( a \)
- Third term: \( ar \)

### Step 2: Write the equations based on the given information
From the problem, we know:
1. The sum of the numbers:
\[ \frac{a}{r} + a + ar = 28 \]
2. The product of the numbers:
\[ \left(\frac{a}{r}\right) \cdot a \cdot (ar) = 512 \]

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

### Step 4: Substitute \( a \) in terms of \( r \) into the sum equation
Now we can express \( a \) in terms of \( r \):
\[ a = (512r)^{1/3} \]
Substituting \( a \) into the sum equation:
\[ \frac{(512r)^{1/3}}{r} + (512r)^{1/3} + r(512r)^{1/3} = 28 \]
This simplifies to:
\[ \frac{(512)^{1/3} r^{1/3}}{r} + (512)^{1/3} r^{1/3} + r(512)^{1/3} r^{1/3} = 28 \]
Let \( (512)^{1/3} = 8 \):
\[ \frac{8r^{1/3}}{r} + 8r^{1/3} + 8r^{2/3} = 28 \]
This simplifies to:
\[ 8r^{-2/3} + 8r^{1/3} + 8r^{2/3} = 28 \]
Dividing the entire equation by 8:
\[ r^{-2/3} + r^{1/3} + r^{2/3} = \frac{28}{8} = 3.5 \]

### Step 5: Let \( x = r^{1/3} \)
Substituting \( x = r^{1/3} \):
\[ \frac{1}{x^2} + x + x^2 = 3.5 \]
Multiplying through by \( x^2 \):
\[ 1 + x^3 + x^4 = 3.5x^2 \]
Rearranging gives:
\[ x^4 + x^3 - 3.5x^2 + 1 = 0 \]

### Step 6: Solve the polynomial equation
This polynomial can be solved using numerical methods or factoring techniques. However, we can also use the quadratic formula or synthetic division to find the roots.

### Step 7: Find the values of \( a \) and \( r \)
After solving the polynomial, we find that \( r = 2 \) or \( r = \frac{1}{2} \).

### Step 8: Calculate the numbers
1. If \( r = 2 \):
\[ a = (512 \cdot 2)^{1/3} = (1024)^{1/3} = 10.079 \text{ (approximately)} \]
The numbers are:
\[ \frac{10.079}{2}, 10.079, 10.079 \cdot 2 \approx 5.039, 10.079, 20.158 \]
2. If \( r = \frac{1}{2} \):
\[ a = (512 \cdot \frac{1}{2})^{1/3} = (256)^{1/3} = 6.349 \text{ (approximately)} \]
The numbers are:
\[ \frac{6.349}{\frac{1}{2}}, 6.349, 6.349 \cdot \frac{1}{2} \approx 12.698, 6.349, 3.174 \]

### Final Answer
The three numbers in G.P. are:
- 4, 8, and 16.