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.

Answer 1

The first term of the geometric series is given as \( a = 5 \) and the common ratio is \( r = 3 \).

The formula for the \( n \)-th term of a geometric series is:

\[
a_n = a \cdot r^{n-1}
\]

Using this formula, we can find the third through seventh terms of the series:

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

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

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

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

5. **Seventh term \( (n = 7) \)**:
\[
a_7 = 5 \cdot 3^{7-1} = 5 \cdot 3^6 = 5 \cdot 729 = 3645
\]

Next, we sum the third through the seventh terms:

\[
\text{Sum} = a_3 + a_4 + a_5 + a_6 + a_7
\]

Calculating the sum:

\[
\text{Sum} = 45 + 135 + 405 + 1215 + 3645
\]

Now performing the addition step by step:

- Sum of \( 45 + 135 = 180 \)
- Sum of \( 180 + 405 = 585 \)
- Sum of \( 585 + 1215 = 1800 \)
- Sum of \( 1800 + 3645 = 6445 \)

Therefore, the sum of the third through the seventh terms of the geometric series is:

\[
\boxed{6445}
\]