Arithmetic Sequences and Series Lesson

In this lesson, we will learn about arithmetic sequences and series. A sequence in which each term after the first is obtained by adding some fixed number to the previous term is known as an arithmetic sequence or an arithmetic progression. The fixed number is known as the common difference and is usually expressed with a lowercase d. $$2, 7, 12, 17, 22,...$$ For example, the above sequence is an arithmetic sequence. Notice that our first term (a₁) is 2. Then each term after is found by adding 5 to the previous term. $$a_1 = 2$$ $$a_2 = 2 + {\color{green} 5} = 7$$ $$a_3 = 7 + {\color{green} 5} = 12$$ $$a_4 = 12 + {\color{green} 5} = 17$$ $$a_5 = 17 + {\color{green} 5} = 22$$

Finding the Common Difference

To find the common difference for an arithmetic sequence, we can use the following formula: $$d=a_{n + 1}- a_{n}$$ In other words, we will choose any two consecutive terms. Then we will subtract the first term (leftmost) from the second term (rightmost). Let's look at an example.
Example #1: Find the common difference. $$29, 39, 49, 59,...$$ We can choose any two consecutive terms. In other words, pick two terms that are next to each other. Since the one on the right has the higher index value, it will serve as the a_{n + 1}, while the one on the left will serve as the a_n. Let's choose a₁ and a₂, which gives us 29 and 39, respectively. We plug into our formula: $$d=a_{n + 1}- a_{n}$$ $$d=a_2 - a_1$$ $$d=39 - 29=10$$ Our common difference (d) is 10. Notice that we could pick any other two consecutive terms and the result would be the same. For example, suppose we instead pick a₃ = 49 and a₄ = 59. $$d=a_{n + 1}- a_{n}$$ $$d=a_4 - a_3$$ $$d = 59 - 49 = 10$$

Finding Terms When Given a₁ and d

If you are given the first term a₁ and the common difference (d), then we can find the first few terms of the sequence pretty easily. The idea is to write the first term a₁, which is given, and then keep adding the common difference (d) to obtain additional terms. Let's look at an example.
Example #2: Find the first five terms of the sequence. The first term (a₁) is 38, and the common difference (d) is (-2). $$a_1 = 38$$ $$a_2 = 38 + {\color{green} (-2) } = 36$$ $$a_3 = 36 + {\color{green} (-2) } = 34$$ $$a_4 = 34 + {\color{green} (-2) } = 32$$ $$a_5 = 32 + {\color{green} (-2) } = 30$$

Finding the n^th Term of an Arithmetic Sequence

In some cases, we will be asked to find the n^th term of an arithmetic sequence and give the explicit formula for a_n. To develop our formula, let's consider the following: $$a_1 = a_1$$ $$a_2 = a_1 + d$$ $$a_3 = a_2 + d = a_1 + d + d = a_1 + 2d$$ $$a_4 = a_3 + d = a_1 + 2d + d = a_1 + 3d$$ If we continue this pattern: $$a_5 = a_1 + 4d$$ $$a_6 = a_1 + 5d$$ Which leads to the following formula. $$a_{n}=a_{1}+ (n - 1)d$$ Let's look at some examples.
Example #3: Find a₂₂ and a_n for the arithmetic sequence. $$28, 31, 34, 37,...$$ In this case, a₁ = 28, and we can find our common difference (d) using our formula. $$d = a_{n + 1} - a_n$$ $$d = a_2 - a_1 = 31 - 28 = 3$$ Now that we have a₁ = 28 and d = 3, we can find a₂₂ with our formula. $$a_{n}=a_{1}+ d(n - 1)$$ $$a_{22}=28 + 3(22 - 1)$$ $$a_{22}=28 + 3 \cdot 21$$ $$a_{22}=28 + 63$$ $$a_{22}=91$$ How do we find the formula for the general term a_n? We just plug in for a₁ and d: $$a_{n}=a_{1}+ d(n - 1)$$ $$a_{n}=28 + 3(n - 1)$$ $$a_{n}=28 + 3n - 3$$ $$a_{n}=25 + 3n$$ Example #4: Find a₃₂ and a_n for the arithmetic sequence. $$a_{10} = -51$$ $$a_{36} = -207$$ Here, we are given two random terms, a₁₀ and a₃₆, that are not consecutive terms. There are different ways of working this type of problem. Using a simple system of equations is probably the most straightforward approach. $$a_{n}=a_{1}+ d(n - 1)$$ Let's get our first equation: $$a_{10} = a_{1} + d(10 - 1)$$ $$a_{10} = a_{1} + 9d$$ $$-51 = a_{1} + 9d$$ Now we can get our second equation: $$a_{36} = a_{1} + d(36 - 1)$$ $$a_{36} = a_{1} + 35d$$ $$-207 = a_{1} + 35d$$ Now we have our system of equations: $$1) \, {-}51 = a_{1} + 9d$$ $$2) \, {-}207 = a_{1} + 35d$$ Solve equation #1 for a₁: $$a_{1} = -9d - 51$$ Plug in for a₁ in equation #2: $$-207 = (-9d - 51) + 35d$$ $$-207 = -9d + 35d - 51$$ $$-156 = 26d$$ $$d = -6$$ Now we can plug -6 in for d in equation #1: $$-51 = a_{1} + 9(-6)$$ $$-51 = a_{1} - 54$$ $$a_{1} = 3$$ Now we can find a₃₂ by plugging into our formula. $$a_{n}=a_{1}+ d(n - 1)$$ $$a_{32} = 3 - 6(32 - 1)$$ $$a_{32} = 3 - 6(31)$$ $$a_{32} = 3 - 186$$ $$a_{32} = -183$$ And to finish things up, let's find a_n by plugging into our formula. $$a_{n} = a_{1}+ d(n - 1)$$ $$a_{n} = 3 - 6(n - 1)$$ $$a_{n} = 3 - 6n + 6$$ $$a_{n} = 9 - 6n$$

Sum of the First n Terms of an Arithmetic Sequence

Recall that a series is the sum of the terms of a sequence. When we sum the terms of an arithmetic sequence, this is known as an arithmetic series. We have two very useful formulas that allow us to find the sum of the first n terms of an arithmetic sequence.
Starting with the sum of the first n terms: $$S_{n} = a_{1} + [a_{1} + d] + [a_{1} + 2d] + \cdots + [a_{1} + (n - 1)d]$$ Next, we will write the same sum S_n in the reverse order. This means we will now start with a_n. Since we are going in reverse, each additional term is found by subtracting d. $$S_{n} = a_{n} + [a_{n} - d] + [a_{n} - 2d] + \cdots + [a_{n} - (n - 1)d]$$ If we add the left sides together and set this equal to the sum of the right sides: $$S_{n} + S_{n} = (a_{1} + a_{n}) + (a_{1} + a_{n}) + \cdots + (a_1 + a_n)$$ We have n terms of (a₁ + a_n) on the right. $$2S_{n} = n(a_{1} + a_{n})$$ Solve for S_n: $$S_{n}=\frac{n}{2}(a_{1}+ a_{n})$$ If you plug in for a_n using the formula from earlier, we can write the result in a different way. $$a_{n} = a_{1}+ d(n - 1)$$ $$S_{n}=\frac{n}{2}[a_{1}+ a_{1} + d(n - 1)]$$ $$S_{n}=\frac{n}{2}[2a_{1}+ (n - 1)d]$$ The first formula will be used when the first and last terms are known. In other cases, the second formula is used. Let's look at a few examples.
Example #5: Evaluate each arithmetic series. $$a_{1} = 2, d = 7, n = 40$$ Since the last term (a₄₀) is not given, let's use the second formula. $$S_{n}=\frac{n}{2}[2a_{1}+ (n - 1)d]$$ $$S_{40}=\frac{40}{2}[2(2) + 7(40 - 1)]$$ $$S_{40}=20[4 + 7(39)]$$ $$S_{40}=20(277)$$ $$S_{40}=5540$$

Using Summation Notation

In the last lesson, we explored the summation properties and rules. Using the formulas we learned in this lesson, we can evaluate an arithmetic series using a different approach. Let's look at an example.
Example #6: Evaluate each series. $$\sum_{i = 1}^{10} (3i + 9)$$ Of course, we can use the strategy from the last lesson. $$\sum_{i = 1}^{10} (3i + 9) = 3\sum_{i = 1}^{10} i + \sum_{i = 1}^{10} 9$$ Scratch Work: $$\sum_{i = 1}^{n} i = \frac{n(n + 1)}{2}$$ $$\sum_{i = 1}^{10} i = \frac{10(10 + 1)}{2} = 55$$ $$\sum_{i = 1}^{n} c = cn$$ $$\sum_{i = 1}^{10} 9 = 9(10) = 90$$ Update: $$3\sum_{i = 1}^{10} i + \sum_{i = 1}^{10} 9 = 3(55) + 90 = 255$$ Let's work the problem using our formula from this lesson. $$a_{1} = 3(1) + 9 = 12$$ $$a_{10} = 3(10) + 9 = 39$$ $$S_{n} = \frac{n}{2}(a_{1} + a_{n})$$ $$S_{10} = \frac{10}{2}(12 + 39) = 5(51) = 255$$

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