A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728, find its first term.

Answer

To solve the problem step by step, we will use the properties of geometric progression (GP) and the formulas for the nth term and the sum of the first n terms.

### Step 1: Identify the given values
We are given:
- Common ratio \( r = 3 \)
- Last term \( t_n = 486 \)
- Sum of the terms \( S_n = 728 \)

### Step 2: Use the formula for the nth term of a GP
The nth term of a geometric progression is given by:
\[ t_n = a \cdot r^{n-1} \]
Where:
- \( a \) is the first term
- \( r \) is the common ratio
- \( n \) is the number of terms

Substituting the known values:
\[ 486 = a \cdot 3^{n-1} \]
This can be rearranged to:
\[ a \cdot 3^{n-1} = 486 \quad \text{(Equation 1)} \]

### Step 3: Use the formula for the sum of the first n terms of a GP
The sum of the first n terms of a geometric progression is given by:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
Substituting the known values:
\[ 728 = \frac{a(3^n - 1)}{3 - 1} \]
This simplifies to:
\[ 728 = \frac{a(3^n - 1)}{2} \]
Multiplying both sides by 2:
\[ 1456 = a(3^n - 1) \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
From Equation 1:
\[ a = \frac{486}{3^{n-1}} \]
Substituting this expression for \( a \) into Equation 2:
\[ 1456 = \left(\frac{486}{3^{n-1}}\right)(3^n - 1) \]
Multiplying both sides by \( 3^{n-1} \):
\[ 1456 \cdot 3^{n-1} = 486(3^n - 1) \]
Expanding the right side:
\[ 1456 \cdot 3^{n-1} = 486 \cdot 3^n - 486 \]
Rearranging gives:
\[ 1456 \cdot 3^{n-1} + 486 = 486 \cdot 3^n \]
Factoring out \( 3^{n-1} \) on the left side:
\[ 1456 + 486 \cdot 3^{-1} = 486 \cdot 3^{1} \]
This simplifies to:
\[ 1456 + \frac{486}{3} = 486 \cdot 3 \]
Calculating \( \frac{486}{3} \):
\[ \frac{486}{3} = 162 \]
Thus:
\[ 1456 + 162 = 1456 + 162 = 486 \cdot 3 \]
Calculating \( 486 \cdot 3 \):
\[ 486 \cdot 3 = 1458 \]

### Step 5: Substitute back to find \( a \)
Now we can substitute \( n \) back into Equation 1 to find \( a \):
Using \( 3^{n-1} = \frac{486}{a} \) and substituting into \( 1456 = a(3^n - 1) \):
\[ 1456 = a(3 \cdot 3^{n-1} - 1) \]
Substituting \( 3^{n-1} = \frac{486}{a} \):
\[ 1456 = a(3 \cdot \frac{486}{a} - 1) \]
This simplifies to:
\[ 1456 = 1458 - a \]
Thus:
\[ a = 1458 - 1456 = 2 \]

### Final Answer
The first term \( a \) is \( 2 \).

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