Algebra 2 Worksheets | Sequences and Series Worksheets - Free Printable

Problem Analysis:

The task involves solving for missing terms in both arithmetic sequences and geometric sequences. Let's break it down step by step.

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Part 1: Arithmetic Sequences

An arithmetic sequence is defined as a sequence where the difference between consecutive terms is constant. This difference is called the common difference ($d$).

The general formula for the $n$-th term of an arithmetic sequence is:
$$
a_n = a_1 + (n-1)d
$$
where:
- $a_n$ is the $n$-th term,
- $a_1$ is the first term,
- $d$ is the common difference,
- $n$ is the term number.

#### Problem 1:
Find the missing term(s) in the sequence: `..., 5, __, -5, ...`

1. Identify the given terms: $a_2 = 5$ and $a_4 = -5$.
2. Let the missing term be $a_3$. The common difference is $d$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 + d = 5$
- For $a_4$: $a_4 = a_1 + 3d = -5$
4. Solve the system of equations:
$$
a_1 + d = 5 \quad \text{(1)}
$$
$$
a_1 + 3d = -5 \quad \text{(2)}
$$
5. Subtract equation (1) from equation (2):
$$
(a_1 + 3d) - (a_1 + d) = -5 - 5
$$
$$
2d = -10 \implies d = -5
$$
6. Substitute $d = -5$ into equation (1):
$$
a_1 + (-5) = 5 \implies a_1 = 10
$$
7. Find $a_3$:
$$
a_3 = a_1 + 2d = 10 + 2(-5) = 10 - 10 = 0
$$

Answer for Problem 1: The missing term is $\boxed{0}$.

---

#### Problem 2:
Find the missing term(s) in the sequence: `..., 4, __, -6, ...`

1. Identify the given terms: $a_2 = 4$ and $a_4 = -6$.
2. Let the missing term be $a_3$. The common difference is $d$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 + d = 4$
- For $a_4$: $a_4 = a_1 + 3d = -6$
4. Solve the system of equations:
$$
a_1 + d = 4 \quad \text{(1)}
$$
$$
a_1 + 3d = -6 \quad \text{(2)}
$$
5. Subtract equation (1) from equation (2):
$$
(a_1 + 3d) - (a_1 + d) = -6 - 4
$$
$$
2d = -10 \implies d = -5
$$
6. Substitute $d = -5$ into equation (1):
$$
a_1 + (-5) = 4 \implies a_1 = 9
$$
7. Find $a_3$:
$$
a_3 = a_1 + 2d = 9 + 2(-5) = 9 - 10 = -1
$$

Answer for Problem 2: The missing term is $\boxed{-1}$.

---

#### Problem 3:
Find the missing term(s) in the sequence: `..., 3, __, __, __, 23, ...`

1. Identify the given terms: $a_2 = 3$ and $a_6 = 23$.
2. Let the missing terms be $a_3$, $a_4$, and $a_5$. The common difference is $d$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 + d = 3$
- For $a_6$: $a_6 = a_1 + 5d = 23$
4. Solve the system of equations:
$$
a_1 + d = 3 \quad \text{(1)}
$$
$$
a_1 + 5d = 23 \quad \text{(2)}
$$
5. Subtract equation (1) from equation (2):
$$
(a_1 + 5d) - (a_1 + d) = 23 - 3
$$
$$
4d = 20 \implies d = 5
$$
6. Substitute $d = 5$ into equation (1):
$$
a_1 + 5 = 3 \implies a_1 = -2
$$
7. Find $a_3$, $a_4$, and $a_5$:
$$
a_3 = a_1 + 2d = -2 + 2(5) = -2 + 10 = 8
$$
$$
a_4 = a_1 + 3d = -2 + 3(5) = -2 + 15 = 13
$$
$$
a_5 = a_1 + 4d = -2 + 4(5) = -2 + 20 = 18
$$

Answer for Problem 3: The missing terms are $\boxed{8, 13, 18}$.

---

#### Problem 4:
Find the missing term(s) in the sequence: `..., 1, __, __, -17, ...`

1. Identify the given terms: $a_2 = 1$ and $a_5 = -17$.
2. Let the missing terms be $a_3$ and $a_4$. The common difference is $d$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 + d = 1$
- For $a_5$: $a_5 = a_1 + 4d = -17$
4. Solve the system of equations:
$$
a_1 + d = 1 \quad \text{(1)}
$$
$$
a_1 + 4d = -17 \quad \text{(2)}
$$
5. Subtract equation (1) from equation (2):
$$
(a_1 + 4d) - (a_1 + d) = -17 - 1
$$
$$
3d = -18 \implies d = -6
$$
6. Substitute $d = -6$ into equation (1):
$$
a_1 + (-6) = 1 \implies a_1 = 7
$$
7. Find $a_3$ and $a_4$:
$$
a_3 = a_1 + 2d = 7 + 2(-6) = 7 - 12 = -5
$$
$$
a_4 = a_1 + 3d = 7 + 3(-6) = 7 - 18 = -11
$$

Answer for Problem 4: The missing terms are $\boxed{-5, -11}$.

---

#### Problem 5:
Find the missing term(s) in the sequence: `..., 2, __, __, 20, ...`

1. Identify the given terms: $a_2 = 2$ and $a_5 = 20$.
2. Let the missing terms be $a_3$ and $a_4$. The common difference is $d$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 + d = 2$
- For $a_5$: $a_5 = a_1 + 4d = 20$
4. Solve the system of equations:
$$
a_1 + d = 2 \quad \text{(1)}
$$
$$
a_1 + 4d = 20 \quad \text{(2)}
$$
5. Subtract equation (1) from equation (2):
$$
(a_1 + 4d) - (a_1 + d) = 20 - 2
$$
$$
3d = 18 \implies d = 6
$$
6. Substitute $d = 6$ into equation (1):
$$
a_1 + 6 = 2 \implies a_1 = -4
$$
7. Find $a_3$ and $a_4$:
$$
a_3 = a_1 + 2d = -4 + 2(6) = -4 + 12 = 8
$$
$$
a_4 = a_1 + 3d = -4 + 3(6) = -4 + 18 = 14
$$

Answer for Problem 5: The missing terms are $\boxed{8, 14}$.

---

#### Problem 6:
Find the missing term(s) in the sequence: `..., 4, __, __, -14, ...`

1. Identify the given terms: $a_2 = 4$ and $a_5 = -14$.
2. Let the missing terms be $a_3$ and $a_4$. The common difference is $d$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 + d = 4$
- For $a_5$: $a_5 = a_1 + 4d = -14$
4. Solve the system of equations:
$$
a_1 + d = 4 \quad \text{(1)}
$$
$$
a_1 + 4d = -14 \quad \text{(2)}
$$
5. Subtract equation (1) from equation (2):
$$
(a_1 + 4d) - (a_1 + d) = -14 - 4
$$
$$
3d = -18 \implies d = -6
$$
6. Substitute $d = -6$ into equation (1):
$$
a_1 + (-6) = 4 \implies a_1 = 10
$$
7. Find $a_3$ and $a_4$:
$$
a_3 = a_1 + 2d = 10 + 2(-6) = 10 - 12 = -2
$$
$$
a_4 = a_1 + 3d = 10 + 3(-6) = 10 - 18 = -8
$$

Answer for Problem 6: The missing terms are $\boxed{-2, -8}$.

---

Part 2: Geometric Sequences

A geometric sequence is defined as a sequence where the ratio between consecutive terms is constant. This ratio is called the common ratio ($r$).

The general formula for the $n$-th term of a geometric sequence is:
$$
a_n = a_1 \cdot r^{n-1}
$$
where:
- $a_n$ is the $n$-th term,
- $a_1$ is the first term,
- $r$ is the common ratio,
- $n$ is the term number.

#### Problem 7:
Find the missing term(s) in the sequence: `..., 1, __, __, __, 16, ...`

1. Identify the given terms: $a_2 = 1$ and $a_6 = 16$.
2. Let the missing terms be $a_3$, $a_4$, and $a_5$. The common ratio is $r$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 \cdot r = 1$
- For $a_6$: $a_6 = a_1 \cdot r^5 = 16$
4. Solve the system of equations:
$$
a_1 \cdot r = 1 \quad \text{(1)}
$$
$$
a_1 \cdot r^5 = 16 \quad \text{(2)}
$$
5. Divide equation (2) by equation (1):
$$
\frac{a_1 \cdot r^5}{a_1 \cdot r} = \frac{16}{1}
$$
$$
r^4 = 16 \implies r = 2 \quad (\text{since } r > 0)
$$
6. Substitute $r = 2$ into equation (1):
$$
a_1 \cdot 2 = 1 \implies a_1 = \frac{1}{2}
$$
7. Find $a_3$, $a_4$, and $a_5$:
$$
a_3 = a_1 \cdot r^2 = \frac{1}{2} \cdot 2^2 = \frac{1}{2} \cdot 4 = 2
$$
$$
a_4 = a_1 \cdot r^3 = \frac{1}{2} \cdot 2^3 = \frac{1}{2} \cdot 8 = 4
$$
$$
a_5 = a_1 \cdot r^4 = \frac{1}{2} \cdot 2^4 = \frac{1}{2} \cdot 16 = 8
$$

Answer for Problem 7: The missing terms are $\boxed{2, 4, 8}$.

---

#### Problem 8:
Find the missing term(s) in the sequence: `..., 1, __, __, __, 256, ...`

1. Identify the given terms: $a_2 = 1$ and $a_6 = 256$.
2. Let the missing terms be $a_3$, $a_4$, and $a_5$. The common ratio is $r$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 \cdot r = 1$
- For $a_6$: $a_6 = a_1 \cdot r^5 = 256$
4. Solve the system of equations:
$$
a_1 \cdot r = 1 \quad \text{(1)}
$$
$$
a_1 \cdot r^5 = 256 \quad \text{(2)}
$$
5. Divide equation (2) by equation (1):
$$
\frac{a_1 \cdot r^5}{a_1 \cdot r} = \frac{256}{1}
$$
$$
r^4 = 256 \implies r = 4 \quad (\text{since } r > 0)
$$
6. Substitute $r = 4$ into equation (1):
$$
a_1 \cdot 4 = 1 \implies a_1 = \frac{1}{4}
$$
7. Find $a_3$, $a_4$, and $a_5$:
$$
a_3 = a_1 \cdot r^2 = \frac{1}{4} \cdot 4^2 = \frac{1}{4} \cdot 16 = 4
$$
$$
a_4 = a_1 \cdot r^3 = \frac{1}{4} \cdot 4^3 = \frac{1}{4} \cdot 64 = 16
$$
$$
a_5 = a_1 \cdot r^4 = \frac{1}{4} \cdot 4^4 = \frac{1}{4} \cdot 256 = 64
$$

Answer for Problem 8: The missing terms are $\boxed{4, 16, 64}$.

---

#### Problem 9:
Find the missing term(s) in the sequence: `..., 5, __, 80, ...`

1. Identify the given terms: $a_2 = 5$ and $a_4 = 80$.
2. Let the missing term be $a_3$. The common ratio is $r$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 \cdot r = 5$
- For $a_4$: $a_4 = a_1 \cdot r^3 = 80$
4. Solve the system of equations:
$$
a_1 \cdot r = 5 \quad \text{(1)}
$$
$$
a_1 \cdot r^3 = 80 \quad \text{(2)}
$$
5. Divide equation (2) by equation (1):
$$
\frac{a_1 \cdot r^3}{a_1 \cdot r} = \frac{80}{5}
$$
$$
r^2 = 16 \implies r = 4 \quad (\text{since } r > 0)
$$
6. Substitute $r = 4$ into equation (1):
$$
a_1 \cdot 4 = 5 \implies a_1 = \frac{5}{4}
$$
7. Find $a_3$:
$$
a_3 = a_1 \cdot r^2 = \frac{5}{4} \cdot 4^2 = \frac{5}{4} \cdot 16 = 20
$$

Answer for Problem 9: The missing term is $\boxed{20}$.

---

#### Problem 10:
Find the missing term(s) in the sequence: `..., 3, __, 108, ...`

1. Identify the given terms: $a_2 = 3$ and $a_4 = 108$.
2. Let the missing term be $a_3$. The common ratio is $r$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 \cdot r = 3$
- For $a_4$: $a_4 = a_1 \cdot r^3 = 108$
4. Solve the system of equations:
$$
a_1 \cdot r = 3 \quad \text{(1)}
$$
$$
a_1 \cdot r^3 = 108 \quad \text{(2)}
$$
5. Divide equation (2) by equation (1):
$$
\frac{a_1 \cdot r^3}{a_1 \cdot r} = \frac{108}{3}
$$
$$
r^2 = 36 \implies r = 6 \quad (\text{since } r > 0)
$$
6. Substitute $r = 6$ into equation (1):
$$
a_1 \cdot 6 = 3 \implies a_1 = \frac{1}{2}
$$
7. Find $a_3$:
$$
a_3 = a_1 \cdot r^2 = \frac{1}{2} \cdot 6^2 = \frac{1}{2} \cdot 36 = 18
$$

Answer for Problem 10: The missing term is $\boxed{18}$.

---

#### Problem 11:
Find the missing term(s) in the sequence: `..., 5, __, __, __, 3125, ...`

1. Identify the given terms: $a_2 = 5$ and $a_6 = 3125$.
2. Let the missing terms be $a_3$, $a_4$, and $a_5$. The common ratio is $r$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 \cdot r = 5$
- For $a_6$: $a_6 = a_1 \cdot r^5 = 3125$
4. Solve the system of equations:
$$
a_1 \cdot r = 5 \quad \text{(1)}
$$
$$
a_1 \cdot r^5 = 3125 \quad \text{(2)}
$$
5. Divide equation (2) by equation (1):
$$
\frac{a_1 \cdot r^5}{a_1 \cdot r} = \frac{3125}{5}
$$
$$
r^4 = 625 \implies r = 5 \quad (\text{since } r > 0)
$$
6. Substitute $r = 5$ into equation (1):
$$
a_1 \cdot 5 = 5 \implies a_1 = 1
$$
7. Find $a_3$, $a_4$, and $a_5$:
$$
a_3 = a_1 \cdot r^2 = 1 \cdot 5^2 = 1 \cdot 25 = 25
$$
$$
a_4 = a_1 \cdot r^3 = 1 \cdot 5^3 = 1 \cdot 125 = 125
$$
$$
a_5 = a_1 \cdot r^4 = 1 \cdot 5^4 = 1 \cdot 625 = 625
$$

Answer for Problem 11: The missing terms are $\boxed{25, 125, 625}$.

---

#### Problem 12:
Find the missing term(s) in the sequence: `..., 2, __, __, 54, ...`

1. Identify the given terms: $a_2 = 2$ and $a_5 = 54$.
2. Let the missing terms be $a_3$ and $a_4$. The common ratio is $r$.
3. Use the formula for the $n$-th term:
- For $a_2$: $a_2 = a_1 \cdot r = 2$
- For $a_5$: $a_5 = a_1 \cdot r^4 = 54$
4. Solve the system of equations:
$$
a_1 \cdot r = 2 \quad \text{(1)}
$$
$$
a_1 \cdot r^4 = 54 \quad \text{(2)}
$$
5. Divide equation (2) by equation (1):
$$
\frac{a_1 \cdot r^4}{a_1 \cdot r} = \frac{54}{2}
$$
$$
r^3 = 27 \implies r = 3 \quad (\text{since } r > 0)
$$
6. Substitute $r = 3$ into equation (1):
$$
a_1 \cdot 3 = 2 \implies a_1 = \frac{2}{3}
$$
7. Find $a_3$ and $a_4$:
$$
a_3 = a_1 \cdot r^2 = \frac{2}{3} \cdot 3^2 = \frac{2}{3} \cdot 9 = 6
$$
$$
a_4 = a_1 \cdot r^3 = \frac{2}{3} \cdot 3^3 = \frac{2}{3} \cdot 27 = 18
$$

Answer for Problem 12: The missing terms are $\boxed{6, 18}$.

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Final Answers:

1. $\boxed{0}$
2. $\boxed{-1}$
3. $\boxed{8, 13, 18}$
4. $\boxed{-5, -11}$
5. $\boxed{8, 14}$
6. $\boxed{-2, -8}$
7. $\boxed{2, 4, 8}$
8. $\boxed{4, 16, 64}$
9. $\boxed{20}$
10. $\boxed{18}$
11. $\boxed{25, 125, 625}$
12. $\boxed{6, 18}$

$\boxed{\text{All problems solved.}}$