Sequences: Videos & Practice Problems

Sequences are ordered lists of numbers where each term can be derived from a pattern. Arithmetic sequences have a constant difference, while geometric sequences have a constant ratio. The general formula for an arithmetic sequence is $a=a{1}_{}+d(n-1)$, and for geometric sequences, it is $a=a{1}_{}r(n-1)$. Understanding these formulas allows for efficient calculation of terms without needing previous values.

Introduction to Sequences

Memorizing numbers, such as bank account numbers or phone numbers, often requires understanding their order. This concept is foundational to sequences, which are essentially lists of numbers arranged in a specific order. For instance, the sequence 2, 4, 6, 8 illustrates how each number is a term, also known as an element or member of the sequence. Recognizing patterns within sequences is crucial; in the example of 2, 4, 6, 8, each term increases by 2. To find the next term, simply add 2 to the last term, resulting in 10.

Sequences can be classified as finite or infinite. A finite sequence has a specific endpoint, while an infinite sequence continues indefinitely. For example, the sequence 3, 6, 9, 12, 18 is finite because it ends at 18, whereas a sequence represented as 1, 1/2, 1, ... continues without end, indicated by the ellipsis.

Sequences share similarities with functions, as both follow specific rules and can be expressed with equations. However, the key difference lies in their inputs. In functions, inputs can be any real number, while in sequences, inputs are called indexes and are always positive integers starting from 1. For example, in the sequence defined by the equation \( a_n = 2n \), the index \( n \) takes on values like 1, 2, 3, etc., producing outputs (terms) such as 2, 4, 6, 8, and so forth.

To illustrate how to find terms in a sequence, consider the formula \( a_n = n^2 \). By substituting the index values into the equation, we can determine the first three terms: for \( n = 1 \), \( a_1 = 1^2 = 1 \); for \( n = 2 \), \( a_2 = 2^2 = 4 \); and for \( n = 3 \), \( a_3 = 3^2 = 9 \). Similarly, for the sequence defined by \( a_n = \frac{1}{n} + 3 \), the first three terms can be calculated as follows: \( a_1 = \frac{1}{1} + 3 = 4 \), \( a_2 = \frac{1}{2} + 3 = 3.5 \), and \( a_3 = \frac{1}{3} + 3 \approx 3.33 \).

Another example involves the sequence defined by \( a_n = (-1)^n \). The first three terms are \( a_1 = (-1)^1 = -1 \), \( a_2 = (-1)^2 = 1 \), and \( a_3 = (-1)^3 = -1 \). This sequence alternates between -1 and 1, demonstrating that sequences can also include negative numbers and exhibit various patterns.

Understanding sequences is essential for recognizing numerical patterns and applying mathematical concepts effectively. With practice, identifying terms and their relationships within sequences becomes a valuable skill in mathematics.

The first 4 terms of a sequence are $\left\lbrace\sqrt3,2\sqrt3,3\sqrt3,4\sqrt3,\ldots\right\rbrace$. Continuing this pattern, find the $7^{\th}$ term.

• A. $8\sqrt3$
• B. $6\sqrt3$
• C. $7\sqrt3$
• D. $9\sqrt3$

Answer: (C)

Determine the first 3 terms of the sequence given by the general formula
$a_{n}=\frac{1}{n!+1}$

• A. $\left\lbrace\frac12,\frac13,\frac17\right\rbrace$
• B. $\left\lbrace\frac12,\frac13,\frac14\right\rbrace$
• C. $\left\lbrace1,2,7\right\rbrace$
• D. $\left\lbrace1,\frac12,\frac16\right\rbrace$

Answer: (A)

Writing a General Formula

Understanding sequences involves recognizing patterns and deriving explicit formulas that represent these patterns mathematically. An explicit formula is a function of the term index \( n \) that allows you to calculate the term directly. The key to finding these formulas lies in identifying the relationships between the terms in the sequence.

One common type of sequence is an arithmetic sequence, where the terms increase by a constant amount. For example, in the sequence \( 5, 6, 7, 8, 9 \), each term increases by 1. The general formula for such sequences can be expressed as:

\[ a_n = n + c \]

where \( c \) is a constant that adjusts the starting point of the sequence. In this case, since the first term is 5, we find that \( c = 4 \), leading to the formula:

\[ a_n = n + 4 \]

Another type of sequence is one that alternates in sign, such as \( -5, 5, -5, 5 \). This pattern can be represented using the expression \( (-1)^n \), which alternates between -1 and 1 based on whether \( n \) is odd or even. To adjust for the specific values in the sequence, we multiply by 5:

\[ a_n = (-1)^n \cdot 5 \]

Sequences can also involve fractions, such as \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6} \). In this case, the numerators follow the pattern \( n \), while the denominators increase by 1 starting from 2. Thus, the general formula becomes:

\[ a_n = \frac{n}{n + 1} \]

Exponential sequences, like \( 2, 4, 8, 16, 32 \), grow at an increasing rate. The general formula for an exponential sequence can be expressed as:

\[ a_n = b^n \]

In this example, since the base is 2, the formula is:

\[ a_n = 2^n \]

By identifying these patterns and applying the appropriate adjustments, you can derive the explicit formulas for various types of sequences. This skill is essential for solving problems related to sequences and series in mathematics.

Writing a General Formula Example 1

In this example, we explore how to derive a general formula for a sequence based on its given terms. The sequence provided consists of fractions: 1/(1×2), 1/(2×3), 1/(3×4), 1/(4×5). To find the general term, denoted as a_n, we analyze the patterns in both the numerators and denominators of these fractions.

First, we observe that the numerator is consistently 1 across all terms, indicating that it does not depend on the term index n. Therefore, the numerator in our general formula will simply be 1.

Next, we examine the denominator, which consists of products of two consecutive integers: 1×2, 2×3, 3×4, 4×5. The first part of the denominator corresponds directly to the term index n, while the second part is always one greater than n. This leads us to conclude that the denominator can be expressed as n(n + 1).

Thus, the general formula for the sequence can be written as:

a_n = \frac{1}{n(n + 1)}

To find a specific term in the sequence, we can substitute a value for n. For example, to find the 15th term, we substitute n = 15 into our formula:

a_{15} = \frac{1}{15(15 + 1)} = \frac{1}{15 \times 16}

Calculating this gives:

a_{15} = \frac{1}{240} \approx 0.0041667

This process illustrates how to derive a general formula from a sequence and use it to calculate specific terms efficiently.

Writing a General Formula Example 2

In this sequence analysis, we explore the pattern of the numbers: -2, 4, -6, 8, -10, and so forth. The goal is to derive a general formula for the nth term, which allows for efficient calculation of any term in the sequence without manually following the pattern. This is particularly useful for determining high-index terms, such as the 18th term.

First, we observe the sequence's behavior. The numbers alternate in sign, and the absolute values increase by 2 with each term. The alternating signs suggest the use of (-1) raised to the power of n, which will yield a negative result for odd n and a positive result for even n. To account for the increasing absolute values, we recognize that the sequence increases by 2 for each term, which can be represented as 2n.

Combining these observations, the general formula for the nth term can be expressed as:

a_n = (-1)ⁿ × 2n

To validate this formula, we can calculate the first few terms:

• For n = 1: a₁ = (-1)¹ × 2 × 1 = -2
• For n = 2: a₂ = (-1)² × 2 × 2 = 4
• For n = 3: a₃ = (-1)³ × 2 × 3 = -6

These calculations confirm that the formula accurately represents the sequence. To find the 18th term, we substitute n = 18 into the formula:

a₁₈ = (-1)¹⁸ × 2 × 18

Since (-1)¹⁸ = 1 (an even power), the calculation simplifies to:

a₁₈ = 1 × 36 = 36

Thus, the 18th term of the sequence is 36. This method of deriving a general formula not only streamlines the process of finding specific terms but also enhances understanding of the underlying patterns in sequences.

Recursive Formulas

In mathematics, sequences can be defined using different types of formulas, primarily general formulas and recursive formulas. A general formula, such as \( a_n = 2n \), allows you to calculate the terms of a sequence directly by substituting the index \( n \). For example, using this formula, the first five terms of the sequence would be 2, 4, 6, 8, and 10 when \( n \) takes on the values 1 through 5.

On the other hand, a recursive formula defines each term based on the previous term(s) in the sequence. For instance, a recursive formula might be expressed as \( a_n = a_{n-1} + 2 \). Here, \( a_{n-1} \) represents the term before \( a_n \). To find the terms of the sequence using this recursive formula, you start with a known initial term, such as \( a_1 = 2 \). From there, you can calculate subsequent terms: \( a_2 = a_1 + 2 = 2 + 2 = 4 \), \( a_3 = a_2 + 2 = 4 + 2 = 6 \), and so forth, resulting in the same sequence of 2, 4, 6, 8, and 10.

The key distinction between general and recursive formulas lies in their approach: general formulas require the index \( n \) to compute the \( n \)-th term directly, while recursive formulas rely on previously calculated terms to derive the next term. This makes recursive formulas particularly useful when the pattern of the sequence is evident, allowing for easier computation of subsequent terms without needing to derive a general formula.

For example, consider a recursive formula defined as \( a_n = 2 \cdot a_{n-1} + 3 \) with an initial term \( a_1 = 1 \). To find the next terms, you would calculate:

• \( a_2 = 2 \cdot a_1 + 3 = 2 \cdot 1 + 3 = 5 \)
• \( a_3 = 2 \cdot a_2 + 3 = 2 \cdot 5 + 3 = 13 \)
• \( a_4 = 2 \cdot a_3 + 3 = 2 \cdot 13 + 3 = 29 \)

Thus, the sequence generated is 1, 5, 13, and 29.

In another example, if you have a recursive formula \( a_n = n \cdot a_{n-1} \) with \( a_1 = 1 \), you would find:

• \( a_2 = 2 \cdot a_1 = 2 \cdot 1 = 2 \)
• \( a_3 = 3 \cdot a_2 = 3 \cdot 2 = 6 \)
• \( a_4 = 4 \cdot a_3 = 4 \cdot 6 = 24 \)

This results in the sequence 1, 2, 6, and 24.

In summary, both general and recursive formulas are valuable tools for defining sequences, each with its own advantages depending on the context and the information available. Understanding how to utilize both types of formulas enhances your ability to work with sequences effectively.

Write the first 6 terms of the sequence given by the recursive formula $a_{n}=a_{n-2}+a_{n-1}$ ; $a_1=1$ ; $a_2=1$.

• A. $\left\lbrace1,1,2,3,5,18\right\rbrace$
• B. $\left\lbrace1,2,3,5,8,13\right\rbrace$
• C. $\left\lbrace0,1,1,2,3,5\right\rbrace$
• D. $\left\lbrace1,1,2,3,5,8\right\rbrace$

Answer: (D)

Arithmetic Sequences - Recursive Formula

Arithmetic sequences are a specific type of sequence where the difference between consecutive terms remains constant. This constant difference is referred to as the common difference, denoted by the letter d. For example, in the sequence 2, 7, 12, 17, each term increases by 5, making d = 5.

To express an arithmetic sequence using a recursive formula, the next term can be calculated by adding the common difference to the previous term. The general form of the recursive formula is:

\( a_n = a_{n-1} + d \)

In this case, for the sequence starting with 2, the recursive formula would be:

\( a_n = a_{n-1} + 5 \)

To illustrate this, if we start with \( a_1 = 2 \), we can find subsequent terms as follows:

• \( a_2 = a_1 + 5 = 2 + 5 = 7 \)
• \( a_3 = a_2 + 5 = 7 + 5 = 12 \)
• \( a_4 = a_3 + 5 = 12 + 5 = 17 \)

Thus, the first four terms of the sequence are 2, 7, 12, and 17.

Sometimes, the common difference can be negative. For instance, if we start with \( a_1 = 9 \) and the recursive formula is \( a_n = a_{n-1} - 6 \), we can calculate the terms as follows:

• \( a_2 = a_1 - 6 = 9 - 6 = 3 \)
• \( a_3 = a_2 - 6 = 3 - 6 = -3 \)
• \( a_4 = a_3 - 6 = -3 - 6 = -9 \)

In this case, the terms are 9, 3, -3, and -9, demonstrating that the common difference can indeed be negative.

When tasked with writing a recursive formula from a given arithmetic sequence, the first step is to determine the common difference by subtracting any two consecutive terms. For example, in the sequence 2, 5, 8, 11, 14, the common difference can be calculated as:

\( d = 5 - 2 = 3 \)

Thus, the recursive formula for this sequence is:

\( a_n = a_{n-1} + 3 \)

To complete the recursive formula, it is essential to specify the initial term, which in this case is \( a_1 = 2 \). Therefore, the complete recursive formula is:

\( a_n = a_{n-1} + 3 \) with \( a_1 = 2 \)

By using these principles, one can effectively work with arithmetic sequences and their recursive representations.

Write a recursive formula for the arithmetic sequence.
$\left\lbrace8,2,-4,-10,\ldots\right\rbrace$

• A. $a_{n}=a_{n-1}-10$ ; $a_1=6$
• B. $a_{n}=a_{n-1}-6$ ; $a_1=6$
• C. $a_{n}=a_{n-1}-6$ ; $a_1=8$
• D. $a_{n}=a_{n-1}-10$ ; $a_1=8$

Answer: (C)

Arithmetic Sequences - General Formula

In arithmetic sequences, understanding how to derive both recursive and general formulas is essential for efficiently calculating terms. A recursive formula defines each term based on the previous term, while a general formula allows for direct calculation of any term in the sequence without needing prior terms.

The general formula for the nth term of an arithmetic sequence can be expressed as:

\[ a_n = a_1 + d \cdot (n - 1) \]

In this formula, a_n represents the nth term, a₁ is the first term, d is the common difference between consecutive terms, and n is the term's index.

To illustrate, consider the sequence 2, 7, 12, 17. Here, the first term a₁ is 2, and the common difference d is 5 (since 7 - 2 = 5). Plugging these values into the general formula gives:

\[ a_n = 2 + 5 \cdot (n - 1) \]

This formula allows for quick calculations of any term. For example, to find the 4th term (a₄), substitute n = 4:

\[ a_4 = 2 + 5 \cdot (4 - 1) = 2 + 15 = 17 \]

This confirms that the 4th term is indeed 17, as expected.

Now, let’s apply this to another sequence: 2, 5, 8, 11, 14. Here, a₁ is 2 and the common difference d is 3. The general formula becomes:

\[ a_n = 2 + 3 \cdot (n - 1) \]

To find the 101st term (a₁₀₁), substitute n = 101:

\[ a_{101} = 2 + 3 \cdot (101 - 1) = 2 + 3 \cdot 100 = 2 + 300 = 302 \]

This demonstrates the efficiency of using the general formula, as calculating the 101st term recursively would be impractical.

In summary, mastering the general formula for arithmetic sequences not only simplifies calculations but also enhances understanding of the sequence's structure and behavior.

Find the general formula for the arithmetic sequence below. Without using a recursive formula, calculate the $30^{\th}$ term.
$\left\lbrace-9,-4,1,6,\ldots\right\rbrace$

• A. 46
• B. 136
• C. 146
• D. 150

Answer: (B)

Arithmetic Sequences - General Formula Example 1

In mathematics, particularly in sequences, converting a recursive formula into a general formula can significantly simplify calculations. For an arithmetic sequence, where each term is derived from the previous term by adding a constant, this process is straightforward.

Consider a recursive formula where the next term is the previous term plus a common difference, denoted as d. For example, if the starting term (first term) is 2 and the common difference is 3, the recursive relationship can be expressed as:

Recursive Formula: a_n = a_n-1 + d, where a₁ = 2 and d = 3.

To derive the general formula for the nth term of the sequence, the formula can be expressed as:

General Formula: a_n = a₁ + d(n - 1).

In this case, substituting the known values gives:

a_n = 2 + 3(n - 1).

This formula allows for the calculation of any term in the sequence without needing to compute all previous terms. For instance, to find the 15th term, substitute n = 15 into the general formula:

a₁₅ = 2 + 3(15 - 1).

Calculating this yields:

a₁₅ = 2 + 3(14) = 2 + 42 = 44.

Thus, the 15th term of the sequence is 44. This method of deriving a general formula from a recursive one is a powerful tool in sequence analysis, allowing for quick calculations of any term in the sequence.

Arithmetic Sequences - General Formula Example 2

In this problem, we are tasked with finding the general term of a sequence given two specific terms: \( a_4 = -2 \) and \( a_6 = 6 \). To approach this, we utilize the general formula for an arithmetic sequence, which is expressed as:

\( a_n = a_1 + d(n - 1) \)

Here, \( a_1 \) represents the first term of the sequence, and \( d \) is the common difference between consecutive terms. Since we are given non-consecutive terms, we need to derive a system of equations to find both \( a_1 \) and \( d \).

Substituting \( n = 4 \) into the general formula gives us:

\( a_4 = a_1 + d(4 - 1) \Rightarrow -2 = a_1 + 3d \quad (1) \

For \( n = 6 \), we have:

\( a_6 = a_1 + d(6 - 1) \Rightarrow 6 = a_1 + 5d \quad (2) \

Now we have a system of two equations:

1. \( a_1 + 3d = -2 \)

2. \( a_1 + 5d = 6 \)

To solve this system, we can use the elimination method. By subtracting equation (1) from equation (2), we eliminate \( a_1 \):

\( (a_1 + 5d) - (a_1 + 3d) = 6 - (-2) \)

This simplifies to:

\( 2d = 8 \)

Dividing both sides by 2 gives us:

\( d = 4 \)

Now that we have the common difference \( d \), we can substitute it back into either equation to find \( a_1 \). Using equation (1):

\( a_1 + 3(4) = -2 \)

This simplifies to:

\( a_1 + 12 = -2 \)

Subtracting 12 from both sides yields:

\( a_1 = -14 \)

With both \( a_1 \) and \( d \) determined, we can now write the general formula for the sequence:

\( a_n = -14 + 4(n - 1) \)

To find a specific term, such as \( a_{18} \), we substitute \( n = 18 \) into the general formula:

\( a_{18} = -14 + 4(18 - 1) \)

This simplifies to:

\( a_{18} = -14 + 4 \times 17 \)

Calculating this gives:

\( a_{18} = -14 + 68 = 54 \)

Thus, the 18th term of the sequence is \( 54 \). This method illustrates how to derive a general formula from known terms in an arithmetic sequence and solve for specific terms using algebraic techniques.

Geometric Sequences - Recursive Formula

In mathematics, sequences can be categorized into different types based on how their terms are generated. One such type is the arithmetic sequence, where each term is derived by adding a constant value, known as the common difference, to the previous term. For example, in the sequence 3, 6, 9, 12, the common difference is 3.

In contrast, a geometric sequence is characterized by a constant ratio between consecutive terms, referred to as the common ratio. For instance, in the sequence 3, 9, 27, 81, each term is obtained by multiplying the previous term by 3. Here, the common ratio (denoted as r) is equal to 3. This fundamental difference in operation—addition for arithmetic sequences and multiplication for geometric sequences—leads to distinct growth patterns. Geometric sequences tend to grow exponentially, resulting in much larger values compared to the linear growth of arithmetic sequences.

To express a geometric sequence recursively, one can use the formula:

\( a_n = a_{n-1} \cdot r \)

In this formula, \( a_n \) represents the current term, \( a_{n-1} \) is the previous term, and \( r \) is the common ratio. For example, if we consider the sequence 5, 20, 80, 320, we can determine the common ratio by dividing any two consecutive terms. Here, from 5 to 20, we multiply by 4, and this pattern continues for the subsequent terms, establishing that the common ratio \( r \) is 4.

Thus, the recursive formula for this sequence can be written as:

\( a_n = a_{n-1} \cdot 4 \)

Additionally, it is essential to specify the first term of the sequence, which in this case is \( a_1 = 5 \). This recursive approach allows for the generation of subsequent terms by simply multiplying the previous term by the common ratio.

Understanding the differences between arithmetic and geometric sequences, along with their respective recursive formulas, is crucial for analyzing patterns in mathematics and applying these concepts in various problem-solving scenarios.

Write a recursive formula for the geometric sequence $\left\lbrace18,6,2,\frac23,\ldots\right\rbrace$.

• A. $a_{n}=3a_{n-1}$
• B. $a_{n}=\frac{a_{n-1}}{3}$
• C. $a_{n}=18a_{n-1}$
• D. $a_{n}=\frac23a_{n-1}$

Answer: (B)

Geometric Sequences - General Formula

In mathematics, understanding sequences is crucial, particularly when distinguishing between arithmetic and geometric sequences. A geometric sequence is defined by a constant ratio between consecutive terms, known as the common ratio (r). To find any term in a geometric sequence without relying on previous terms, we use a general formula.

The general formula for the nth term of a geometric sequence can be expressed as:

\( a_n = a_1 \cdot r^{n-1} \)

Here, \( a_n \) represents the nth term, \( a_1 \) is the first term of the sequence, and \( r \) is the common ratio. This formula allows us to calculate any term directly, which is particularly useful for finding terms at high indices.

For example, consider the geometric sequence 3, 6, 12, 24. The first term \( a_1 \) is 3, and the common ratio \( r \) is 2 (since each term is multiplied by 2 to get the next term). Using the general formula, we can express the nth term as:

\( a_n = 3 \cdot 2^{n-1} \)

To find the fourth term (\( n = 4 \)), we substitute into the formula:

\( a_4 = 3 \cdot 2^{4-1} = 3 \cdot 2^3 = 3 \cdot 8 = 24 \)

This confirms that the fourth term is indeed 24.

Now, let’s apply this to another sequence: 5, 20, 80, 320. Here, the first term \( a_1 \) is 5, and the common ratio \( r \) is 4. The general formula for this sequence becomes:

\( a_n = 5 \cdot 4^{n-1} \)

To find the twelfth term (\( n = 12 \)), we substitute into the formula:

\( a_{12} = 5 \cdot 4^{12-1} = 5 \cdot 4^{11} \)

Calculating \( 4^{11} \) yields a large number, specifically 4,194,304. Thus, the twelfth term is:

\( a_{12} = 5 \cdot 4,194,304 = 20,971,520 \)

This example illustrates the power of the general formula in efficiently calculating terms in a geometric sequence, especially when dealing with high indices. Understanding these concepts is essential for mastering sequences and series in mathematics.

Find the $10^{\th}$ term of the geometric sequence in which $a_1=5$ and $r=2$.

• A. 5120
• B. 1280
• C. 10240
• D. 2560

Answer: (D)

Write a formula for the general or $n^{\th}$ term of the geometric sequence where $a_7=1458$ and $r=-3$.

• A. $a_{n}=1\cdot\left(-3\right)^{n-1}$
• B. $a_{n}=2\cdot\left(-3\right)^{n-1}$
• C. $a_{n}=-\frac23\cdot\left(-3\right)^{n-1}$
• D. $a_{n}=\frac23\cdot\left(-3\right)^{n-1}$

Answer: (B)

Here’s what students ask on this topic:

What is the difference between arithmetic and geometric sequences?

Arithmetic and geometric sequences differ in how their terms are generated. In an arithmetic sequence, the difference between consecutive terms is constant, called the common difference (d). For example, in the sequence 2, 5, 8, 11, the common difference is 3. The general formula for an arithmetic sequence is:

$a_n=a_1+d(n-1)$

In a geometric sequence, each term is obtained by multiplying the previous term by a constant ratio (r). For example, in the sequence 3, 9, 27, 81, the common ratio is 3. The general formula for a geometric sequence is:

$a_n=a_1\times r^{n-1}$

Understanding these differences is crucial for solving problems involving sequences.

How do you find the nth term of an arithmetic sequence?

To find the nth term of an arithmetic sequence, use the general formula:

$a_n=a_1+d(n-1)$

Here:

• $a_1$ is the first term of the sequence.
• $d$ is the common difference between terms.
• $n$ is the position of the term you want to find.

For example, in the sequence 3, 7, 11, 15, where $a_1$ = 3 and $d$ = 4, the 5th term is:

$a_5=3+4(5-1)=19$

Thus, the 5th term is 19.

What is a recursive formula for a sequence?

A recursive formula defines each term of a sequence based on the previous term(s). It is particularly useful for sequences where the relationship between consecutive terms is straightforward. For example, in an arithmetic sequence with a common difference $d$, the recursive formula is:

$a_n=a_{n-1}+d$

For geometric sequences with a common ratio $r$, the recursive formula is:

$a_n=a_{n-1}\times r$

Additionally, the first term $a_1$ must be specified. For example, if $a_1$ = 2 and $d$ = 3, the sequence is 2, 5, 8, 11, etc.

How do you write a general formula for a geometric sequence?

To write a general formula for a geometric sequence, use the equation:

$a_n=a_1\times r^{n-1}$

Here:

• $a_1$ is the first term of the sequence.
• $r$ is the common ratio.
• $n$ is the position of the term you want to find.

For example, in the sequence 2, 6, 18, 54, where $a_1$ = 2 and $r$ = 3, the general formula is:

$a_n=2\times 3^{n-1}$

Using this formula, you can calculate any term in the sequence efficiently.

What is the common difference in an arithmetic sequence?

The common difference in an arithmetic sequence is the constant amount added to or subtracted from each term to get the next term. It is denoted by $d$. To find the common difference, subtract any term from the subsequent term:

$d=a_{n+1}-a_n$

For example, in the sequence 4, 7, 10, 13, the common difference is:

$d=7-4=3$

Thus, $d$ = 3. This value is crucial for writing formulas and analyzing the sequence.