# Exploring Sequence and Series Patterns
A sequence is an ordered list of numbers, while a series is the sum of the elements of a sequence. Both sequences and series are fundamental concepts in mathematics, especially in calculus, algebra, and number theory. Understanding the patterns in sequences and series can be incredibly useful for solving problems related to them.
## Arithmetic and Geometric Sequences
The two most common sequence patterns are arithmetic and geometric sequences. An **arithmetic sequence** is one in which each term is obtained by adding a constant, called the common difference, to the previous term. The common difference can be positive, negative, or zero. For example, in the arithmetic sequence 2, 4, 6, 8, …, the common difference is 2.
A **geometric sequence** is one where each term is obtained by multiplying the previous term by a constant called the common ratio. For example, in the geometric sequence 3, 6, 12, 24, …, the common ratio is 2.
### Arithmetic Sequence Formula:
If \(a_1\) is the first term and \(d\) is the common difference, then the \(n\)th term of an arithmetic sequence (\(a_n\)) is given by:
\[ a_n = a_1 + (n – 1)d \]
### Arithmetic Series Formula:
The sum of the first \(n\) terms of an arithmetic series \(S_n\) can be found by:
\[ S_n = \frac{n}{2}(2a_1 + (n – 1)d) \]
### Geometric Sequence Formula:
If \(a_1\) is the first term and \(r\) is the common ratio, then the \(n\)th term of a geometric sequence (\(a_n\)) is given by:
\[ a_n = a_1 \cdot r^{(n – 1)} \]
### Geometric Series Formula:
The sum of the first \(n\) terms of a geometric series \(S_n\) is:
\[ S_n = \frac{a_1(1 – r^n)}{1 – r} \]
(for \(r \neq 1\))
Understanding these formulas is crucial for solving problems related to sequence and series patterns.
## Problems and Solutions
Here are 20 problems on sequence and series patterns along with their solutions using the concepts discussed above.
### Arithmetic Sequence Problems
1. Find the 10th term of the arithmetic sequence: 5, 9, 13, 17, …
*Solution*: Using the formula for the nth term of an arithmetic sequence,
\(a_n = a_1 + (n – 1)d\),
\(a_{10} = 5 + (10 – 1) \cdot 4 = 5 + 36 = 41\).
2. Determine the 15th term of the arithmetic sequence if the first term is 8 and the common difference is 3.
*Solution*: \(a_{15} = 8 + (15 – 1) \cdot 3 = 8 + 42 = 50\).
3. Find the sum of the first 20 terms of the arithmetic series: 2, 5, 8, …
*Solution*: Using the formula for the sum of an arithmetic series,
\(S_n = \frac{n}{2}(2a_1 + (n – 1)d)\),
\(S_{20} = \frac{20}{2}(2 \cdot 2 + (20 – 1) \cdot 3) = 10(4 + 57) = 610\).
4. The 5th term of an arithmetic sequence is 18, and the common difference is 2. Find the 1st term.
*Solution*: \(a_5 = a_1 + 4d\). Thus, \(18 = a_1 + 4 \cdot 2\), so \(a_1 = 10\).
5. Calculate the 12th term of an arithmetic sequence with the 7th term as 23 and the common difference of -3.
*Solution*: \(a_{12} = a_7 + (12 – 7)(-3) = 23 + (5)(-3) = 8\).
### Geometric Sequence Problems
6. Find the 4th term in the geometric sequence: 3, 6, 12, …
*Solution*: \(a_4 = a_1 \cdot r^{(4 – 1)} = 3 \cdot 2^{3} = 3 \cdot 8 = 24\).
7. What is the 8th term of a geometric sequence where the first term is 16 and the common ratio is 1/2?
*Solution*: \(a_8 = 16 \cdot (1/2)^{7} = 16 \cdot 1/128 = 1/8\).
8. Calculate the sum of the first 5 terms of the geometric series: 32, 64, 128, …
*Solution*: \(S_n = \frac{a_1(1 – r^n)}{1 – r}\),
\(S_5 = \frac{32(1 – 2^5)}{1 – 2} = \frac{32(-31)}{-1} = 992\).
9. If the 3rd term of a geometric sequence is 27 and the common ratio is 3, find the 1st term.
*Solution*: \(a_3 = a_1 \cdot r^{2}\). Solving for \(a_1\), we get \(a_1 = \frac{27}{3^{2}} = 3\).
10. What is the sum of the first 7 terms of the geometric series with the first term of 5 and a common ratio of -2?
*Solution*: \(S_7 = \frac{5(1 – (-2)^7)}{1 – (-2)} = \frac{5(1 + 128)}{3} = \frac{645}{3} = 215\).
### Mixed Arithmetic and Geometric Sequence Problems
11. In an arithmetic sequence, the 3rd term is 15 and the 8th term is 35. Find the common difference.
*Solution*: \(a_3 = a_1 + 2d\) and \(a_8 = a_1 + 7d\). Since \(a_8 – a_3 = 35 – 15 = 20\) and \(7d – 2d = 5d\), we have that \(5d = 20\), thus \(d = 4\).
12. If the 5th term of a geometric sequence is 81 and the 2nd term is 9, find the common ratio.
*Solution*: Since \(a_5 = a_1 \cdot r^{4}\) and \(a_2 = a_1 \cdot r\), it follows that \(r^3 = \frac{a_5}{a_2} = \frac{81}{9} = 9\), hence \(r = 3\).
13. A certain arithmetic series has a sum of 100 for its first 10 terms. If the common difference is 2, find the first term.
*Solution*: Using the formula for the sum we have \(100 = \frac{10}{2}(2a_1 + (10 – 1) \cdot 2)\), which simplifies to \(10a_1 + 90 = 100\), thus \(a_1 = 1\).
14. Find the sum of first \(n\) terms of an arithmetic series whose \(n\)th term is given by \(3n + 4\).
*Solution*: The sum of an arithmetic series with the \(n\)th term \(3n + 4\) is \(S_n = \frac{n}{2}(a_1 + (3n + 4))\). Since \(a_1 = 3(1) + 4 = 7\), then \(S_n = \frac{n}{2}(7+ 3n + 4) = \frac{n}{2}(3n + 11)\).
15. How many terms are in the arithmetic series 5 + 8 + 11 + … + 56?
*Solution*: The common difference is 3, and the last term is 56. So \(a_n = 5 + (n – 1) \cdot 3 = 56\). Solving for \(n\), we get \(n = 18\).
16. If the 4th term of a geometric series is 64 and the 7th term is 512, what is the common ratio?
*Solution*: Using \(a_4 = a_1 \cdot r^{3}\) and \(a_7 = a_1 \cdot r^{6}\), we have \(\frac{512}{64} = r^{3}\), which gives us \(r = 2\).
17. What is the 15th term of the sequence whose nth term is given as \(a_n = 2^n + 1\)?
*Solution*: \(a_{15} = 2^{15} + 1\).
18. Find the 9th term of the arithmetic sequence if \(a_1 = -5\) and \(d = -3\).
*Solution*: \(a_9 = -5 + (9 – 1)(-3) = -5 – 24 = -29\).
19. Calculate the sum of the infinite geometric series: 7, -3.5, 1.75, …
*Solution*: This is an infinite geometric series with \(a_1 = 7\) and \(r = -1/2\). The sum \(S\) is given by \(S = \frac{a_1}{1 – r} = \frac{7}{1 – (-1/2)} = \frac{7}{3/2} = \frac{14}{3}\).
20. If the sum of the first 6 terms of a geometric series is 63 and the first term is 1, what is the common ratio?
*Solution*: Using the formula for the sum, we get \(63 = \frac{1(1 – r^6)}{1 – r}\). We need to find \(r\) that satisfies this equation. By testing common ratios through substitution, if \(r = 2\) then \(S_6 = 63\), so \(r = 2\).
These problems are examples of the type of questions that arise in algebra and calculus courses, and they help to build a deeper understanding of the patterns and behaviors of sequences and series.
