Find three numbers in G.P. whose sum is 19 and product is 216.

Answer

To find three numbers in Geometric Progression (G.P.) whose sum is 19 and product is 216, we can follow these steps:

### Step 1: Define the terms in G.P.
Let the three numbers in G.P. be \( \frac{a}{r}, a, ar \), where \( a \) is the middle term and \( r \) is the common ratio.

### Step 2: Set up the equations
From the problem, we know:
1. The sum of the numbers is 19:
\[ \frac{a}{r} + a + ar = 19 \]
2. The product of the numbers is 216:
\[ \frac{a}{r} \cdot a \cdot ar = 216 \]

### Step 3: Simplify the product equation
The product simplifies to:
\[ \frac{a^3}{r} = 216 \]
Multiplying both sides by \( r \):
\[ a^3 = 216r \]

### Step 4: Solve for \( a \)
Taking the cube root of both sides:
\[ a = \sqrt[3]{216r} = 6\sqrt[3]{r} \]

### Step 5: Substitute \( a \) in the sum equation
Substituting \( a \) back into the sum equation:
\[ \frac{6\sqrt[3]{r}}{r} + 6\sqrt[3]{r} + 6\sqrt[3]{r}r = 19 \]
This simplifies to:
\[ \frac{6}{\sqrt[3]{r^2}} + 6\sqrt[3]{r} + 6r^{\frac{4}{3}} = 19 \]

### Step 6: Clear the fractions
Multiply through by \( r^{\frac{2}{3}} \) to eliminate the fraction:
\[ 6 + 6r + 6r^2 = 19r^{\frac{2}{3}} \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 6r^2 - 19r^{\frac{2}{3}} + 6r = 0 \]

### Step 8: Let \( x = r^{\frac{1}{3}} \)
Let \( x = r^{\frac{1}{3}} \), then \( r = x^3 \):
\[ 6x^6 - 19x^2 + 6x^3 = 0 \]

### Step 9: Solve the cubic equation
This is a polynomial in \( x \). We can use numerical methods or factorization to find the roots.

### Step 10: Find values of \( r \)
Once we find the values of \( r \), we can substitute back to find \( a \) and subsequently the three numbers.

### Step 11: Calculate the numbers
For each value of \( r \), calculate:
1. \( \frac{a}{r} \)
2. \( a \)
3. \( ar \)

### Final Result
The numbers will be either \( 4, 6, 9 \) or \( 9, 6, 4 \).