Three numbers are in G.P. Their sum is 14. If we multiply the first and third numbers by 4 and 2nd number by 5, the new numbers form an A.P. Find the numbers.

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Define the numbers in G.P.
Let the three numbers in geometric progression (G.P.) be represented as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)

### Step 2: Set up the equation for the sum of the numbers
According to the problem, the sum of these three numbers is 14:
\[ a + ar + ar^2 = 14 \]
Factoring out \( a \):
\[ a(1 + r + r^2) = 14 \quad \text{(Equation 1)} \]

### Step 3: Set up the equation for the new numbers forming an A.P.
When we multiply the first number by 4, the second number by 5, and the third number by 2, the new numbers are:
- First number: \( 4a \)
- Second number: \( 5ar \)
- Third number: \( 2ar^2 \)

These new numbers must form an arithmetic progression (A.P.). For numbers \( x, y, z \) to be in A.P., the condition is:
\[ y = \frac{x + z}{2} \]
Applying this to our new numbers:
\[ 5ar = \frac{4a + 2ar^2}{2} \]

### Step 4: Simplify the A.P. condition
Multiplying both sides by 2 to eliminate the fraction:
\[ 10ar = 4a + 2ar^2 \]
Rearranging gives:
\[ 2ar^2 - 10ar + 4a = 0 \]
Factoring out \( 2a \):
\[ 2a(r^2 - 5r + 2) = 0 \]

### Step 5: Solve the quadratic equation
Since \( a \neq 0 \), we can divide by \( 2a \):
\[ r^2 - 5r + 2 = 0 \]
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2} \]

### Step 6: Find values of \( r \)
The roots are:
\[ r_1 = \frac{5 + \sqrt{17}}{2}, \quad r_2 = \frac{5 - \sqrt{17}}{2} \]

### Step 7: Substitute back to find \( a \)
Using Equation 1:
\[ a(1 + r + r^2) = 14 \]
Calculating \( 1 + r + r^2 \) for each \( r \):
1. For \( r_1 \):
\[ r_1^2 = \left(\frac{5 + \sqrt{17}}{2}\right)^2 = \frac{25 + 10\sqrt{17} + 17}{4} = \frac{42 + 10\sqrt{17}}{4} \]
\[ 1 + r_1 + r_1^2 = 1 + \frac{5 + \sqrt{17}}{2} + \frac{42 + 10\sqrt{17}}{4} \]
Simplifying this will give a value for \( a \).

2. For \( r_2 \):
\[ r_2^2 = \left(\frac{5 - \sqrt{17}}{2}\right)^2 = \frac{25 - 10\sqrt{17} + 17}{4} = \frac{42 - 10\sqrt{17}}{4} \]
\[ 1 + r_2 + r_2^2 = 1 + \frac{5 - \sqrt{17}}{2} + \frac{42 - 10\sqrt{17}}{4} \]
Again, simplifying this will give another value for \( a \).

### Step 8: Find the three numbers
Once \( a \) is found for each case of \( r \), the three numbers in G.P. can be computed as:
- For \( r_1 \): \( a, ar_1, ar_1^2 \)
- For \( r_2 \): \( a, ar_2, ar_2^2 \)