To calculate the partial sum \( S_5 \) for the sequence \( 243, 81, 27, \ldots \), we first identify the type of sequence we have.
The given sequence can be expressed as:
- The first term \( a_1 = 243 \)
- The second term \( a_2 = 81 \)
- The third term \( a_3 = 27 \)
We can see that each term is obtained by multiplying the previous term by \( \frac{1}{3} \):
\[ a_2 = a_1 \times \frac{1}{3} = 243 \times \frac{1}{3} = 81 \] \[ a_3 = a_2 \times \frac{1}{3} = 81 \times \frac{1}{3} = 27 \]
This indicates that the sequence is a geometric series where:
- The first term \( a = 243 \)
- The common ratio \( r = \frac{1}{3} \)
The \( n \)-th term of a geometric series can be expressed as:
\[ a_n = a r^{n-1} \]
Now we need to compute the partial sum \( S_5 \) of the first 5 terms:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
Substituting for \( n = 5 \):
\[ S_5 = 243 \frac{1 - \left(\frac{1}{3}\right)^5}{1 - \frac{1}{3}} \]
Calculating \( \left(\frac{1}{3}\right)^5 \):
\[ \left(\frac{1}{3}\right)^5 = \frac{1}{243} \]
Now substituting this back into the formula:
\[ S_5 = 243 \frac{1 - \frac{1}{243}}{\frac{2}{3}} \]
Simplifying the numerator:
\[ 1 - \frac{1}{243} = \frac{243 - 1}{243} = \frac{242}{243} \]
Thus:
\[ S_5 = 243 \cdot \frac{\frac{242}{243}}{\frac{2}{3}} = 243 \cdot \frac{242}{243} \cdot \frac{3}{2} \]
The \( 243 \) cancels out:
\[ S_5 = \frac{242 \cdot 3}{2} = \frac{726}{2} = 363 \]
Therefore, the partial sum \( S_5 \) is \( \boxed{363} \).
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