In a geometric series, the \( n \)-th term can be calculated using the formula:
\[ a_n = a_1 \cdot r^{n-1} \]
where \( a_1 \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.
For this problem:
- The first term \( a_1 = 5 \)
- The common ratio \( r = 3 \)
We need to find the third through seventh terms (i.e., \( a_3, a_4, a_5, a_6, \) and \( a_7 \)) and sum them up.
Let's calculate each term:
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Third term (\( a_3 \)): \[ a_3 = 5 \cdot 3^{3-1} = 5 \cdot 3^2 = 5 \cdot 9 = 45 \]
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Fourth term (\( a_4 \)): \[ a_4 = 5 \cdot 3^{4-1} = 5 \cdot 3^3 = 5 \cdot 27 = 135 \]
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Fifth term (\( a_5 \)): \[ a_5 = 5 \cdot 3^{5-1} = 5 \cdot 3^4 = 5 \cdot 81 = 405 \]
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Sixth term (\( a_6 \)): \[ a_6 = 5 \cdot 3^{6-1} = 5 \cdot 3^5 = 5 \cdot 243 = 1215 \]
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Seventh term (\( a_7 \)): \[ a_7 = 5 \cdot 3^{7-1} = 5 \cdot 3^6 = 5 \cdot 729 = 3645 \]
Now, we sum these terms:
\[ S = a_3 + a_4 + a_5 + a_6 + a_7 \]
Calculating the sum:
\[ S = 45 + 135 + 405 + 1215 + 3645 \]
Calculating step-by-step:
\[ 45 + 135 = 180 \] \[ 180 + 405 = 585 \] \[ 585 + 1215 = 1800 \] \[ 1800 + 3645 = 5445 \]
Thus, the sum of the third through seventh terms of the geometric series is:
\[ \boxed{5445} \]
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