Find the 4 terms in G .P. in which 3rd term is 9 more than the first term and 2nd term is 18 more than the 4th term.

Explanation

To find the four terms in a geometric progression (G.P.) where the third term is 9 more than the first term and the second term is 18 more than the fourth term, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Terms of the G.P.**:
Let the first term be \( a \) and the common ratio be \( r \). The four terms in G.P. can be expressed as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)
- Fourth term: \( ar^3 \)

2. **Set Up the Equations**:
According to the problem:
- The third term is 9 more than the first term:
\[ ar^2 = a + 9 \quad \text{(Equation 1)} \]
- The second term is 18 more than the fourth term:
\[ ar = ar^3 + 18 \quad \text{(Equation 2)} \]

3. **Rearranging Equation 1**:
From Equation 1:
\[ ar^2 - a = 9 \]
Factoring out \( a \):
\[ a(r^2 - 1) = 9 \quad \text{(Equation 3)} \]

4. **Rearranging Equation 2**:
From Equation 2:
\[ ar - ar^3 = 18 \]
Factoring out \( ar \):
\[ ar(1 - r^2) = 18 \quad \text{(Equation 4)} \]

5. **Dividing Equation 4 by Equation 3**:
Now, we can divide Equation 4 by Equation 3:
\[ \frac{ar(1 - r^2)}{a(r^2 - 1)} = \frac{18}{9} \]
Simplifying the right side:
\[ \frac{ar(1 - r^2)}{a(r^2 - 1)} = 2 \]
Since \( r^2 - 1 = -(1 - r^2) \), we have:
\[ -r = 2 \quad \Rightarrow \quad r = -2 \]

6. **Finding the Value of \( a \)**:
Substitute \( r = -2 \) back into Equation 3:
\[ a((-2)^2 - 1) = 9 \]
Simplifying:
\[ a(4 - 1) = 9 \quad \Rightarrow \quad 3a = 9 \quad \Rightarrow \quad a = 3 \]

7. **Calculating the Four Terms**:
Now we can find the four terms:
- First term: \( a = 3 \)
- Second term: \( ar = 3 \times (-2) = -6 \)
- Third term: \( ar^2 = 3 \times (-2)^2 = 3 \times 4 = 12 \)
- Fourth term: \( ar^3 = 3 \times (-2)^3 = 3 \times (-8) = -24 \)

Thus, the four terms in the G.P. are:
\[ 3, -6, 12, -24 \]

### Final Answer:
The four terms in the G.P. are **3, -6, 12, -24**.