If the fourth and nineth terms of a G.P. are 54 and 13122 respectively, find the G.P. Also find its general term.

Answer

To solve the problem, we need to find the geometric progression (G.P.) given that the fourth term \( T_4 \) is 54 and the ninth term \( T_9 \) is 13122. We will also find the general term of the G.P.

### Step-by-Step Solution:

1. **Understanding the General Term of a G.P.**:
The general term \( T_n \) of a G.P. can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

2. **Setting Up the Equations**:
From the problem, we have:
- \( T_4 = a \cdot r^{4-1} = a \cdot r^3 = 54 \) (Equation 1)
- \( T_9 = a \cdot r^{9-1} = a \cdot r^8 = 13122 \) (Equation 2)

3. **Dividing the Two Equations**:
To eliminate \( a \), we can divide Equation 2 by Equation 1:
\[ \frac{T_9}{T_4} = \frac{a \cdot r^8}{a \cdot r^3} \]
This simplifies to:
\[ \frac{13122}{54} = r^{8-3} = r^5 \]
Now, calculating \( \frac{13122}{54} \):
\[ \frac{13122}{54} = 243 \]
Thus, we have:
\[ r^5 = 243 \]

4. **Finding the Common Ratio \( r \)**:
We can express 243 as a power of 3:
\[ 243 = 3^5 \]
Therefore, taking the fifth root:
\[ r = 3 \]

5. **Finding the First Term \( a \)**:
Now that we have \( r \), we can substitute it back into Equation 1 to find \( a \):
\[ a \cdot r^3 = 54 \]
Substituting \( r = 3 \):
\[ a \cdot 3^3 = 54 \]
\[ a \cdot 27 = 54 \]
Dividing both sides by 27:
\[ a = \frac{54}{27} = 2 \]

6. **Forming the G.P.**:
Now that we have both \( a \) and \( r \):
- First term \( a = 2 \)
- Common ratio \( r = 3 \)

The G.P. can be expressed as:
\[ 2, 6, 18, 54, 162, \ldots \]

7. **Finding the General Term**:
The general term \( T_n \) of the G.P. is:
\[ T_n = a \cdot r^{n-1} = 2 \cdot 3^{n-1} \]

### Final Answer:
The G.P. is \( 2, 6, 18, 54, 162, \ldots \) and the general term is \( T_n = 2 \cdot 3^{n-1} \).