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

Explanation

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 Numbers
Let the three numbers in G.P. be \( a \), \( ar \), and \( ar^2 \), where \( a \) is the first term and \( r \) is the common ratio.

### Step 2: Set Up the Equations
From the problem, we know:
1. The sum of the numbers:
\[ a + ar + ar^2 = 19 \]
This can be factored as:
\[ a(1 + r + r^2) = 19 \tag{1} \]

2. The product of the numbers:
\[ a \cdot ar \cdot ar^2 = 216 \]
This simplifies to:
\[ a^3 r^3 = 216 \tag{2} \]

### Step 3: Solve for \( a \) and \( r \)
From equation (2), we can express \( a \) in terms of \( r \):
\[ a^3 = \frac{216}{r^3} \implies a = \sqrt[3]{\frac{216}{r^3}} = \frac{6}{r} \]

### Step 4: Substitute \( a \) in Equation (1)
Substituting \( a = \frac{6}{r} \) into equation (1):
\[ \frac{6}{r}(1 + r + r^2) = 19 \]
Multiplying through by \( r \) to eliminate the fraction:
\[ 6(1 + r + r^2) = 19r \]
Expanding and rearranging gives:
\[ 6 + 6r + 6r^2 - 19r = 0 \implies 6r^2 - 13r + 6 = 0 \]

### Step 5: Solve the Quadratic Equation
Now we can solve the quadratic equation \( 6r^2 - 13r + 6 = 0 \) using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 6 \cdot 6}}{2 \cdot 6} \]
Calculating the discriminant:
\[ b^2 - 4ac = 169 - 144 = 25 \]
So,
\[ r = \frac{13 \pm 5}{12} \]
Calculating the two possible values for \( r \):
1. \( r = \frac{18}{12} = \frac{3}{2} \)
2. \( r = \frac{8}{12} = \frac{2}{3} \)

### Step 6: Find Corresponding Values of \( a \)
Using \( r = \frac{3}{2} \):
\[ a = \frac{6}{\frac{3}{2}} = 4 \]
The numbers are:
\[ 4, 6, 9 \]

Using \( r = \frac{2}{3} \):
\[ a = \frac{6}{\frac{2}{3}} = 9 \]
The numbers are:
\[ 9, 6, 4 \]

### Conclusion
Thus, the three numbers in G.P. are \( 4, 6, 9 \).

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