Somewhere along the road, you likely encountered a problem of this type on a standardized test:

In the following sequence, which number is missing? {78, 71, 64, 57, ___, 43, 36} A. 51 B. 50 C. 49 D. 48 |

The answer is **Choice B.** To solve such a question, we need to find a pattern by examining the successive terms. It looks as though the rule is “subtract 7.” So, the term that is missing is 57 – 7 = 50.

This question involves a sequence. A **sequence** is a list of numbers in a particular order. Each number in a sequence is called a **term**. A sequence can have a finite number of terms, as in the example above, or an infinite number of terms.

A **series** is the sum of the terms in a sequence. Students often get the terms “sequences” and “series” confused or will use the words interchangeably.

In today’s lesson, we’ll learn more about sequences and series and their importance in math. Let’s get started!

## Arithmetic Sequences

Let’s look at some examples of sequences.

{4, 1, -2, -5, -8, -11}

This sequence is an example of an **arithmetic sequence**. As in our first example, each pair of successive terms differ by a constant. That constant is called the **common difference**.

The common difference in this case is 1 – 4 = -3, found by subtracting the first term from the second term. We can verify this by subtracting another pair of successive terms: -5 – -2 = -3.

We have an arithmetic sequence with a common difference of -3. Note, this sequence is finite; there are 6 terms in the sequence.

An arithmetic sequence is a sequence in which each pair of successive terms differs by the same constant, the common difference, denoted by d. The of an arithmetic sequence is nth termdenoted a._{n}An arithmetic sequence is a linear function whose domain is the positive integers. It is linear because the function has a constant rate of change. |

Here are some typical math problems that involve arithmetic sequences:

### Example 1

Find the next two terms in the arithmetic sequence:

{100, 114, 128, 142, …}

**Solution**

We can easily see that the common difference *d* is 114 – 100 = 14. To determine the next two terms we simply add 14 to the previous term:

*a*_{5}= 142 + 14 = 156*a*_{6}= 156 + 14 = 170

The next two terms are 156 and 170.

### Example 2

Find the formula for *a** _{n}* , the

*n*th term of the arithmetic sequence {-3, -7, -11, -15, -19…}.

**Solution**

To find the formula for the *n*th term of the sequence, we can start by finding the common difference:

-7 – (-3) = -4

We can find the formula by working with the common difference -4 and the first term -3. Because arithmetic sequences relate to linear functions, it seems that we can write an expression in a manner that resembles a linear function. Let’s try this equation:

*a*_{n}*n*

Let’s check that this formula is accurate. From the sequence, we can see that the 5th term is -19. That means if we plug *n* = 5 into our formula, we should obtain -19.

*a*_{n}*n**a*= -3 + -4(5)_{5}*a*= -3 + -20_{5}*a*= -23_{5}

Wait, this isn’t right. The term -23 would be the 6th term, not the 5th. We can simply adjust our formula by subtracting 1 from *n*. Let’s check to see if this formula will work.

*a*_{n}*(n – 1)**a*= -3 + -4(5 – 1)_{5}*a*= -3 + -16_{5}*a*= -19_{5}

Our formula looks good! Let’s generalize this result.

The formula for a_{n}, _{ }the nth term of an arithmetic sequence, is given by:a = _{n}a + _{1}d(n – 1)where a is the first term of the sequence _{1}and d is the common difference. |

As mentioned before, a series is the sum of a sequence. More specifically, an **arithmetic series** is the sum of the terms in an arithmetic sequence.

In general, for arithmetic series, we’ll mainly be working with series that find the sum of a sequence with a finite number of terms.

## Arithmetic Series

### Example 3

Find the sum 6 + 11 + 16 + 21 + …+ 96.

**Solution**

It would be difficult to actually add all these numbers. There must be a faster way, right? Yes there is! The German mathematician Carl Friedrich Gauss figured out a way to quickly determine this sum.

Start by writing the sum, let’s call it *S*.

S = 6 + 11 + 16 + … + 96

Rewrite the sum S in reverse order and line up the terms underneath the original sum. Since the common difference is *d* = 5, we can see that the last few terms will be 86 + 91 + 96.

Now we’ll add the equations.

S = | 6 + 11 + 16 + . . . + 86 + 91 + 96 |

S = | 96 + 91 + 86 + . . . + 16 + 11 + 6 |

2S = | 102 + 102 +102 + … + 102 + 102 + 102 |

Notice, the sum of each column on the right side is 102. To solve for the sum S, we can apply a little algebra.

We need to know the number of 102’s though. How do we figure out the number of terms? Here’s a formula to help:

The number of terms in an arithmetic series is equal to:(Last Term – First Term)/d + 1 where d is the common difference. |

In our series, the number of terms is (96 – 6)/5 + 1 = 19. Now let’s find the sum.

We have 19 terms on the right side with each term equaling 102.

- 2S = (19)(102)
- 2S = 1,938
- S = 1,938/2
- S = 969

The sum is 969!

We can generalize Gauss’ discovery with this formula:

The sum Sis equal to of _{n}n terms in an arithmetic series S = n(a_{n}_{1}+ a_{n})/2where a_{1} is the first term and a is the last term._{n} |

Let’s try this formula using the previous example. Since *n* is the number of terms, we have *n *= 19. The first term *a*_{1} = 6 and the 19th term is 96, so *a*_{19} = 96.

- S
= 19 (6+ 96)2_{n} - S
= 191022_{n} - S
= 969_{n}

Notice our sum of 969 is the same result as when we used Gauss’ method. That’s good news!

Now let’s look at another type of sequence.

## Geometric Sequences

### Example 4

Find the next two terms in the sequence {10, 5, 5/2, 5/4, 5/8, …}.

**Solution**

In this example, we can pretty easily see that the sequence is NOT arithmetic. If we subtract pairs of successive terms, we do not get a common difference.

To determine what the pattern is, we can divide successive terms or set up a ratio of successive terms. Divide the second term by the first term to obtain: 5/10=1/2.

Now, let’s confirm that this is the pattern by setting up the ratio of the fourth term to the third term: (5/4)/(5/2)=1/2. This sequence has a common ratio of ½. It is a **geometric sequence**.

Let’s find the next two terms in the sequence by multiplying by ½.

The sixth term, denoted by *a*_{6 }, is equal to (5/8)*(1/2)=5/16.

The seventh term *a*_{7} is (5/16)*(1/2)=5/32.

Seems easy enough, right? In a **geometric sequence**, the ratio of any pair of successive terms is constant. In our example, the sequence is infinite.

We can continue the pattern by multiplying each term by ½ to determine the next term.

A geometric sequence is a sequence in which the ratio of any pair of successive terms is the same constant. The constant is called the common ratio, denoted by r.A geometric sequence is an exponential function whose domain is the positive integers. |

### Example 5

Find the next two terms of the geometric sequence: {-1, 3, -9, 27, …}.

**Solution**

To determine the next two terms, we first need the common ratio *r*. We find this by dividing any term by the preceding term.

- r=3/-1

- r= -3

The common ratio is -3. To find the *a*_{5 }, the 5th term, we multiply 27 by the common ratio. To find the 6th term, we multiply the 5th term by the common ratio.

*a*_{5 }= (27)(-3) = -81*a*_{6 }= (-81)(-3) = 243

The next two terms in the sequence are -81 and 243.

### Example 6

Find a** **formula for *a*_{n} , the *n*th term of the geometric sequence {12, 4, 4/3, 4/9, …}.

**Solution**

When we developed a formula for the *n*th term of an arithmetic sequence, we worked with the first term and the common difference. Here, it makes sense to once again work with the first term but this time we’ll work with the common ratio.

Our first term is 12 and our common ratio is easily calculated:

- r = 4/12= 1/3

Let’s try this formula, using an exponential model for the expression.

*a*= 12(1/3)_{n}^{n}

Now let’s test that the formula is accurate. We know from the given sequence that the 4th term is 4/9 . Let’s see if the formula will give us this value when we plug in *n* = 4:

*a*= 12(1/3)_{n}^{n}*a*= 12(1/3)4_{4}*a*= 12(1/81)_{4}*a*= 12/81_{4}*a*= 4/27_{4}

Hmmm, 4/27 is not the 4th term; in fact, it’s the 5th term so we need to adjust our formula. As before, we can just subtract 1 from *n* so we’ll get the 4th term.

So our formula becomes:

*a*= 12(1/3)_{n}^{n-1}

If you test our new formula, you’ll see that it works!!!

The formula for a_{n},_{ } the nth term of a geometric sequence, is given by:a = _{n}a(_{1}r)^{n}^{ – 1}where a is the first term of the _{1}sequence and r is the common ratio. |

## Geometric Series

Now, let’s turn our attention to **geometric series**. A geometric series is the sum of a geometric sequence.

Consider the geometric series 1/2+1/4+1/8+1/16+… Notice, this series is infinite!

So, how can we find the sum? In this series, the common ratio is 1/2.

It turns out that if the common ratio of an infinite geometric series is between -1 and 1, the **series converges**, meaning the sum is a finite number!!! Amazing!

For an infinite geometric series with |r|<1, the sum of the series is given by the formula a _{1}/(1-r)where r is the common ratio and a_{1} is the first term. |

### Example 7

Find the sum of the geometric series {64 + 8 + 1 + 1/8+…}

**Solution**

Start by finding the common ratio:

- r = 8/64=1/8

Since the common ratio is less than 1, we can use the formula for the sum of an infinite geometric series. The first term *a*_{1 } is 64.

Plugging into our formula, we obtain:

- Sum = 64/(1 – 1/8)
- Sum = 64/(7/8)
- Sum = 512/7

The sum of the infinite geometric series is 512/7. We say that the series **converges** since there is a finite number for the infinite sum.

Hopefully this result astounds you! We found the sum of an infinite series!

It may at least give you something to think over the course of the day.

Sequences and Series are big topics in math. Now you have a better understanding of the differences between the two.

You can learn about quadratic sequences here.

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**About the author:**

Jean-Marie Gard is an independent math teacher and tutor based in Massachusetts. You can get in touch with Jean-Marie at https://testpreptoday.com/.