Geometric Sequences - Math Worksheets - Free Printable

The task involves finding the number of terms (\( n \)) in each geometric sequence given the first term (\( a \)), common ratio (\( r \)), and the last term (\( l \)). The formula used to solve for \( n \) is:

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

We will solve each problem step by step.

---

Problem 1:

- \( a = 5 \)
- \( r = 3 \)
- \( l = 10935 \)

#### Solution:
1. Use the formula \( l = a \cdot r^{n-1} \):
\[
10935 = 5 \cdot 3^{n-1}
\]

2. Divide both sides by 5:
\[
\frac{10935}{5} = 3^{n-1}
\]
\[
2187 = 3^{n-1}
\]

3. Express 2187 as a power of 3:
\[
2187 = 3^7
\]

4. Equate the exponents:
\[
3^{n-1} = 3^7
\]
\[
n - 1 = 7
\]

5. Solve for \( n \):
\[
n = 7 + 1
\]
\[
n = 8
\]

Answer: \( n = 8 \)

---

Problem 2:

- \( a = 6 \)
- \( r = \frac{1}{3} \)
- \( l = \frac{2}{243} \)

#### Solution:
1. Use the formula \( l = a \cdot r^{n-1} \):
\[
\frac{2}{243} = 6 \cdot \left( \frac{1}{3} \right)^{n-1}
\]

2. Divide both sides by 6:
\[
\frac{\frac{2}{243}}{6} = \left( \frac{1}{3} \right)^{n-1}
\]
\[
\frac{2}{243 \cdot 6} = \left( \frac{1}{3} \right)^{n-1}
\]
\[
\frac{2}{1458} = \left( \frac{1}{3} \right)^{n-1}
\]
\[
\frac{1}{729} = \left( \frac{1}{3} \right)^{n-1}
\]

3. Express \(\frac{1}{729}\) as a power of \(\frac{1}{3}\):
\[
\frac{1}{729} = \left( \frac{1}{3} \right)^6
\]

4. Equate the exponents:
\[
\left( \frac{1}{3} \right)^{n-1} = \left( \frac{1}{3} \right)^6
\]
\[
n - 1 = 6
\]

5. Solve for \( n \):
\[
n = 6 + 1
\]
\[
n = 7
\]

Answer: \( n = 7 \)

---

Problem 3:

- \( a = \sqrt{6} \)
- \( r = \sqrt{2} \)
- \( l = 8\sqrt{3} \)

#### Solution:
1. Use the formula \( l = a \cdot r^{n-1} \):
\[
8\sqrt{3} = \sqrt{6} \cdot (\sqrt{2})^{n-1}
\]

2. Divide both sides by \(\sqrt{6}\):
\[
\frac{8\sqrt{3}}{\sqrt{6}} = (\sqrt{2})^{n-1}
\]

3. Simplify the left side:
\[
\frac{8\sqrt{3}}{\sqrt{6}} = \frac{8\sqrt{3}}{\sqrt{2} \cdot \sqrt{3}} = \frac{8}{\sqrt{2}} = 8 \cdot \frac{\sqrt{2}}{2} = 4\sqrt{2}
\]

4. So, we have:
\[
4\sqrt{2} = (\sqrt{2})^{n-1}
\]

5. Express \( 4\sqrt{2} \) as a power of \(\sqrt{2}\):
\[
4\sqrt{2} = (\sqrt{2})^3
\]

6. Equate the exponents:
\[
(\sqrt{2})^{n-1} = (\sqrt{2})^3
\]
\[
n - 1 = 3
\]

7. Solve for \( n \):
\[
n = 3 + 1
\]
\[
n = 6
\]

Answer: \( n = 6 \)

---

Problem 4:

- \( a = -2 \)
- \( r = -2 \)
- \( l = 16384 \)

#### Solution:
1. Use the formula \( l = a \cdot r^{n-1} \):
\[
16384 = -2 \cdot (-2)^{n-1}
\]

2. Divide both sides by \(-2\):
\[
\frac{16384}{-2} = (-2)^{n-1}
\]
\[
-8192 = (-2)^{n-1}
\]

3. Note that \((-2)^{n-1}\) must be negative, so \( n-1 \) must be odd. Express \(-8192\) as a power of \(-2\):
\[
-8192 = (-2)^{14}
\]

4. Equate the exponents:
\[
(-2)^{n-1} = (-2)^{14}
\]
\[
n - 1 = 14
\]

5. Solve for \( n \):
\[
n = 14 + 1
\]
\[
n = 15
\]

Answer: \( n = 15 \)

---

Problem 5:

- \( a = 144 \)
- \( r = \frac{1}{6} \)
- \( l = \frac{1}{1944} \)

#### Solution:
1. Use the formula \( l = a \cdot r^{n-1} \):
\[
\frac{1}{1944} = 144 \cdot \left( \frac{1}{6} \right)^{n-1}
\]

2. Divide both sides by 144:
\[
\frac{\frac{1}{1944}}{144} = \left( \frac{1}{6} \right)^{n-1}
\]
\[
\frac{1}{1944 \cdot 144} = \left( \frac{1}{6} \right)^{n-1}
\]
\[
\frac{1}{280608} = \left( \frac{1}{6} \right)^{n-1}
\]

3. Express \(\frac{1}{280608}\) as a power of \(\frac{1}{6}\):
\[
280608 = 6^8 \quad \text{(since } 6^8 = 1679616 \text{ and } \frac{1}{280608} = \frac{1}{6^8})
\]
\[
\frac{1}{280608} = \left( \frac{1}{6} \right)^8
\]

4. Equate the exponents:
\[
\left( \frac{1}{6} \right)^{n-1} = \left( \frac{1}{6} \right)^8
\]
\[
n - 1 = 8
\]

5. Solve for \( n \):
\[
n = 8 + 1
\]
\[
n = 9
\]

Answer: \( n = 9 \)

---

Problem 6:

- \( a = 4.5 \)
- \( r = 2 \)
- \( l = 73728 \)

#### Solution:
1. Use the formula \( l = a \cdot r^{n-1} \):
\[
73728 = 4.5 \cdot 2^{n-1}
\]

2. Divide both sides by 4.5:
\[
\frac{73728}{4.5} = 2^{n-1}
\]
\[
16384 = 2^{n-1}
\]

3. Express 16384 as a power of 2:
\[
16384 = 2^{14}
\]

4. Equate the exponents:
\[
2^{n-1} = 2^{14}
\]
\[
n - 1 = 14
\]

5. Solve for \( n \):
\[
n = 14 + 1
\]
\[
n = 15
\]

Answer: \( n = 15 \)

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Final Answers:

\[
\boxed{8, 7, 6, 15, 9, 15}
\]