Three number whose product it 512 are in G.P. If
8 is added to the first and 6 to the second , the
number will be in A.P. The numbers are

Options

A. 2, 8,32
B. 8,8,8
C. 4,8,16
D. 2,8,14

Answer

To solve the problem, we need to find three numbers in geometric progression (G.P.) whose product is 512 and, when 8 is added to the first number and 6 to the second, the numbers will be in arithmetic progression (A.P.).

Let's denote the three numbers in G.P. as:
- First term: \( a/r \)
- Second term: \( a \)
- Third term: \( ar \)

### Step 1: Set up the equation for the product
The product of the three numbers is given as:
\[ \left(\frac{a}{r}\right) \cdot a \cdot (ar) = 512 \]
This simplifies to:
\[ \frac{a^3}{r} = 512 \]
Multiplying both sides by \( r \):
\[ a^3 = 512r \]

### Step 2: Find the value of \( a \)
Taking the cube root of both sides, we find:
\[ a = \sqrt[3]{512r} = 8\sqrt[3]{r} \]
Since we are looking for integer values, we will consider \( r \) such that \( a \) remains an integer.

### Step 3: Set up the equation for A.P.
According to the problem, when we add 8 to the first term and 6 to the second term, the new sequence becomes:
- First term: \( \frac{a}{r} + 8 \)
- Second term: \( a + 6 \)
- Third term: \( ar \)

For these to be in A.P., the condition is:
\[ 2(a + 6) = \left(\frac{a}{r} + 8\right) + (ar) \]
Expanding this gives:
\[ 2a + 12 = \frac{a}{r} + ar + 8 \]

### Step 4: Rearranging the equation
Rearranging the equation:
\[ 2a + 12 - 8 = \frac{a}{r} + ar \]
This simplifies to:
\[ 2a + 4 = \frac{a}{r} + ar \]

### Step 5: Multiply through by \( r \)
To eliminate the fraction, multiply through by \( r \):
\[ r(2a + 4) = a + ar^2 \]
This can be rearranged to:
\[ 2ar + 4r - a - ar^2 = 0 \]
Rearranging gives:
\[ ar^2 - 2ar - 4r + a = 0 \]

### Step 6: Substitute \( a \)
Substituting \( a = 8 \) (from our earlier finding):
\[ 8r^2 - 16r - 4r + 8 = 0 \]
This simplifies to:
\[ 8r^2 - 20r + 8 = 0 \]

### Step 7: Solve the quadratic equation
Dividing the entire equation by 4:
\[ 2r^2 - 5r + 2 = 0 \]
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]
Calculating the discriminant:
\[ r = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \]
This gives us:
\[ r = \frac{8}{4} = 2 \quad \text{or} \quad r = \frac{2}{4} = \frac{1}{2} \]

### Step 8: Find the three numbers
1. If \( r = 2 \):
- First term: \( \frac{8}{2} = 4 \)
- Second term: \( 8 \)
- Third term: \( 8 \cdot 2 = 16 \)

So, the numbers are \( 4, 8, 16 \).

2. If \( r = \frac{1}{2} \):
- First term: \( \frac{8}{\frac{1}{2}} = 16 \)
- Second term: \( 8 \)
- Third term: \( 8 \cdot \frac{1}{2} = 4 \)

So, the numbers are \( 16, 8, 4 \).

### Final Answer
The three numbers are \( 4, 8, 16 \) or \( 16, 8, 4 \) (which are essentially the same set).