Geometric Sequence interactive worksheet - Free Printable

Problem: Solving the Geometric Sequence Worksheet

We are tasked with solving problems related to geometric sequences. The general formula for the \( n \)-th term of a geometric sequence is:

\[
a_n = a_1 r^{n-1}
\]

Where:
- \( a_1 \) is the first term,
- \( r \) is the common ratio,
- \( n \) is the term number,
- \( a_n \) is the \( n \)-th term.

Let's solve each problem step by step.

---

Problem 1:

Find \( a_{11} \) of the geometric sequence \( 3, 6, 12, 24, \ldots \).

#### Step 1: Identify \( a_1 \), \( r \), and \( n \).
- The first term \( a_1 = 3 \).
- The common ratio \( r \) is found by dividing any term by its preceding term:
\[
r = \frac{6}{3} = 2
\]
- We need to find the 11th term, so \( n = 11 \).

#### Step 2: Use the formula \( a_n = a_1 r^{n-1} \).
\[
a_{11} = a_1 r^{11-1} = 3 \cdot 2^{10}
\]

#### Step 3: Calculate \( 2^{10} \).
\[
2^{10} = 1024
\]

#### Step 4: Compute \( a_{11} \).
\[
a_{11} = 3 \cdot 1024 = 3072
\]

#### Final Answer:
\[
\boxed{3072}
\]

---

Problem 2:

Find the 9th term of the geometric sequence \( 4, -12, 36, -108, \ldots \).

#### Step 1: Identify \( a_1 \), \( r \), and \( n \).
- The first term \( a_1 = 4 \).
- The common ratio \( r \) is found by dividing any term by its preceding term:
\[
r = \frac{-12}{4} = -3
\]
- We need to find the 9th term, so \( n = 9 \).

#### Step 2: Use the formula \( a_n = a_1 r^{n-1} \).
\[
a_9 = a_1 r^{9-1} = 4 \cdot (-3)^8
\]

#### Step 3: Calculate \( (-3)^8 \).
Since the exponent is even, \( (-3)^8 \) is positive:
\[
(-3)^8 = 3^8 = 6561
\]

#### Step 4: Compute \( a_9 \).
\[
a_9 = 4 \cdot 6561 = 26244
\]

#### Final Answer:
\[
\boxed{26244}
\]

---

Problem 3:

In the geometric sequence \( \frac{1}{5}, \frac{1}{10}, \frac{1}{20}, \frac{1}{40}, \ldots \), what is \( a_{10} \)?

#### Step 1: Identify \( a_1 \), \( r \), and \( n \).
- The first term \( a_1 = \frac{1}{5} \).
- The common ratio \( r \) is found by dividing any term by its preceding term:
\[
r = \frac{\frac{1}{10}}{\frac{1}{5}} = \frac{1}{10} \cdot \frac{5}{1} = \frac{1}{2}
\]
- We need to find the 10th term, so \( n = 10 \).

#### Step 2: Use the formula \( a_n = a_1 r^{n-1} \).
\[
a_{10} = a_1 r^{10-1} = \frac{1}{5} \cdot \left(\frac{1}{2}\right)^9
\]

#### Step 3: Calculate \( \left(\frac{1}{2}\right)^9 \).
\[
\left(\frac{1}{2}\right)^9 = \frac{1}{2^9} = \frac{1}{512}
\]

#### Step 4: Compute \( a_{10} \).
\[
a_{10} = \frac{1}{5} \cdot \frac{1}{512} = \frac{1}{2560}
\]

#### Final Answer:
\[
\boxed{\frac{1}{2560}}
\]

---

Problem 4:

What term is 2048 in the geometric sequence \( 2, 4, 8, 16, \ldots \)?

#### Step 1: Identify \( a_1 \), \( r \), and \( a_n \).
- The first term \( a_1 = 2 \).
- The common ratio \( r \) is found by dividing any term by its preceding term:
\[
r = \frac{4}{2} = 2
\]
- We need to find the term number \( n \) such that \( a_n = 2048 \).

#### Step 2: Use the formula \( a_n = a_1 r^{n-1} \).
\[
2048 = 2 \cdot 2^{n-1}
\]

#### Step 3: Simplify the equation.
\[
2048 = 2^1 \cdot 2^{n-1} = 2^{n}
\]

#### Step 4: Solve for \( n \).
Since \( 2048 = 2^{11} \):
\[
2^n = 2^{11} \implies n = 11
\]

#### Final Answer:
\[
\boxed{11}
\]

---

Summary of Answers:

1. \( a_{11} = 3072 \)
2. The 9th term is \( 26244 \)
3. \( a_{10} = \frac{1}{2560} \)
4. The term 2048 is the \( 11 \)-th term.

\[
\boxed{3072, 26244, \frac{1}{2560}, 11}
\]