Certainly! To find the sum of the first nine terms of a geometric sequence where the 3rd term is 45 and the 6th term is 1215, let's use the properties and formulas of geometric sequences.
### Step-by-Step Solution:
1. Understand the Given Information:
- The 3rd term (\(\(a_3\)\)) is 45.
- The 6th term (\(\(a_6\)\)) is 1215.
2. Formulate the General Form of the nth Term:
The nth term of a geometric sequence can be expressed as:
\(\[
a_n = a \cdot r^{n-1}
\]\)
where:
- \(\(a\)\) is the first term of the sequence.
- \(\(r\)\) is the common ratio.
3. Set Up Equations Using the Given Terms:
- For the 3rd term (\(\(a_3\)\)):
\(\[
45 = a \cdot r^2
\]\)
- For the 6th term (\(\(a_6\)\)):
\(\[
1215 = a \cdot r^5
\]\)
4. Determine the Common Ratio (r):
Divide the equation for the 6th term by the equation for the 3rd term:
\(\[
\frac{1215}{45} = \frac{a \cdot r^5}{a \cdot r^2}
\]\)
Simplifying the fraction:
\(\[
27 = r^3
\]\)
Solve for \(\(r\)\):
\(\[
r = \sqrt[3]{27} = 3
\]\)
5. Determine the First Term (a):
Use the value of \(\(r\)\) in the equation for the 3rd term:
- Given: \(\(45 = a \cdot 3^2\)\)
- Simplify to find \(\(a\)\):
\(\[
45 = a \cdot 9
\]\)
\(\[
a = \frac{45}{9} = 5
\]\)
6. Find the Sum of the First Nine Terms:
The sum \(\(S_n\)\) of the first \(\(n\)\) terms of a geometric sequence is given by:
\(\[
S_n = a \cdot \frac{1 - r^n}{1 - r} \quad \text{(for } r
eq 1\text{)}
\]\)
Substitute \(\(a = 5\)\), \(\(r = 3\)\), and \(\(n = 9\)\):
\(\[
S_9 = 5 \cdot \frac{1 - 3^9}{1 - 3}
\]\)
Calculate \(\(3^9\)\):
\(\[
3^9 = 19683
\]\)
Substitute back into the sum formula:
\(\[
S_9 = 5 \cdot \frac{1 - 19683}{1 - 3}
\]\)
Calculate the numerator:
\(\[
1 - 19683 = -19682
\]\)
Then the fraction:
\(\[
\frac{-19682}{-2} = 9841
\]\)
Finally, multiply by \(\(a\)\):
\(\[
S_9 = 5 \cdot 9841 = 49205
\]\)
Thus, the sum of the first nine terms of the geometric sequence is \(\(49205\)\).
