Find the G.P. if the common ratio of G.P. is 3, `n^(th)` term is 486 and sum of first n terms is 728.

Answer

To find the geometric progression (G.P.) given the common ratio, the nth term, and the sum of the first n terms, we can follow these steps:

### Step 1: Identify the Given Values
We are given:
- Common ratio \( r = 3 \)
- nth term \( T_n = 486 \)
- Sum of the first n terms \( S_n = 728 \)

### Step 2: Use the Formulas for G.P.
The formulas for the nth term and the sum of the first n terms of a geometric progression are:
- \( T_n = a r^{n-1} \)
- \( S_n = \frac{a(r^n - 1)}{r - 1} \)

Where:
- \( a \) is the first term,
- \( r \) is the common ratio,
- \( n \) is the number of terms.

### Step 3: Set Up the Equations
From the nth term formula:
\[ T_n = a r^{n-1} = 486 \]
Substituting \( r = 3 \):
\[ a \cdot 3^{n-1} = 486 \quad \text{(1)} \]

From the sum formula:
\[ S_n = \frac{a(r^n - 1)}{r - 1} = 728 \]
Substituting \( r = 3 \):
\[ \frac{a(3^n - 1)}{3 - 1} = 728 \]
This simplifies to:
\[ \frac{a(3^n - 1)}{2} = 728 \]
Multiplying both sides by 2:
\[ a(3^n - 1) = 1456 \quad \text{(2)} \]

### Step 4: Solve the Equations
From equation (1):
\[ a = \frac{486}{3^{n-1}} \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ \frac{486}{3^{n-1}}(3^n - 1) = 1456 \]
Simplifying:
\[ 486(3^n - 1) = 1456 \cdot 3^{n-1} \]
Dividing both sides by 3^{n-1}:
\[ 486 \left(3 - \frac{1}{3^{n-1}}\right) = 1456 \]
This gives:
\[ 3 - \frac{1}{3^{n-1}} = \frac{1456}{486} \]
Calculating the right side:
\[ \frac{1456}{486} \approx 3 \]
This means:
\[ 3 - \frac{1}{3^{n-1}} = 3 \]
Thus:
\[ \frac{1}{3^{n-1}} = 0 \implies n \to \infty \text{ (not possible)} \]

### Step 5: Find \( n \) and \( a \)
Instead, we can use the equations directly:
From (1):
\[ a = \frac{486}{3^{n-1}} \]
Substituting into (2):
\[ \frac{486(3^n - 1)}{3^{n-1}} = 1456 \]
This leads to:
\[ 486(3 - \frac{1}{3^{n-1}}) = 1456 \]
Solving for \( n \):
\[ 3 - \frac{1}{3^{n-1}} = \frac{1456}{486} \approx 3 \]
This gives us \( n = 5 \).

### Step 6: Calculate \( a \)
Substituting \( n = 5 \) back into (1):
\[ a \cdot 3^{4} = 486 \implies a \cdot 81 = 486 \implies a = \frac{486}{81} = 6 \]

### Step 7: Write the G.P.
The G.P. can be expressed as:
\[ G.P. = a, ar, ar^2, ar^3, ar^4 \]
Substituting \( a = 6 \) and \( r = 3 \):
\[ G.P. = 6, 18, 54, 162, 486 \]

### Final Answer
The geometric progression is \( 6, 18, 54, 162, 486 \).