Question

Find the sum of the third through seventh terms of the geometric series whose first term is 5 and the common ratio is 3.(1 point)
S=

Answer 1

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:

1. **Third term (\( a_3 \)):**
\[
a_3 = 5 \cdot 3^{3-1} = 5 \cdot 3^2 = 5 \cdot 9 = 45
\]

2. **Fourth term (\( a_4 \)):**
\[
a_4 = 5 \cdot 3^{4-1} = 5 \cdot 3^3 = 5 \cdot 27 = 135
\]

3. **Fifth term (\( a_5 \)):**
\[
a_5 = 5 \cdot 3^{5-1} = 5 \cdot 3^4 = 5 \cdot 81 = 405
\]

4. **Sixth term (\( a_6 \)):**
\[
a_6 = 5 \cdot 3^{6-1} = 5 \cdot 3^5 = 5 \cdot 243 = 1215
\]

5. **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:

\[
\boxed{5445}
\]