Geometric Sequences and Series Lesson

In this lesson, we will learn about geometric sequences and series. A geometric sequence, which is also called a geometric progression, is a sequence where each term after the first is obtained by multiplying the preceding term by a fixed nonzero real number known as the common ratio (r). We will use a lowercase r to represent our common ratio. $$1, 5, 25, 125, 625,...$$ For example, the sequence above is a geometric sequence. The first term is 1, and each term afterwards is found by multiplying the preceding term by 5. $$a_1 = 1$$ $$a_2 = 1 \cdot 5 = 5$$ $$a_3 = 5 \cdot 5 = 25$$ $$a_4 = 25 \cdot 5 = 125$$ $$a_5 = 125 \cdot 5 = 625$$ If we were to divide any term after the first one by the preceding term, we would obtain the common ratio r = 5. $$r = \frac{a_2}{a_1} = \frac{5}{1} = 5$$ $$r = \frac{a_4}{a_3} = \frac{125}{25} = 5$$

Finding the Common Ratio

To find r, our common ratio, we can use the following formula: $$r=\frac{a_{n + 1}}{a_{n}}$$ We can choose any two consecutive terms in our geometric sequence. In other words, these are terms that are next to each other. We will then divide the rightmost term by the leftmost term to get the common ratio. Let's look at an example.
Example #1: Find the common ratio. $$-4, 24, -144, 864,...$$ $$r=\frac{a_{n + 1}}{a_{n}}$$ We can pick any two consecutive terms or terms that are next to each other. Let's use a₁ and a₂. $$r=\frac{a_2}{a_1} = \frac{24}{-4}=-6$$ This would work with any other valid choices. Let's also use a₂ and a₃. $$r = \frac{a_3}{a_2} = \frac{-144}{24} = -6$$

Finding the n^th term

Additionally, we may be asked to find the n^th term of a geometric sequence or to find the formula for the general term a_n. To develop our formula, let's consider the following: $$a_1 = a_1$$ $$a_2 = a_1 \cdot r$$ $$a_3 = a_2 \cdot r = a_1 \cdot r \cdot r = a_1 \cdot r^2$$ $$a_4 = a_3 \cdot r = a_1 \cdot r^2 \cdot r = a_1 \cdot r^3$$ If we continue this pattern: $$a_5 = a_1 \cdot r^4$$ $$a_6 = a_1 \cdot r^5$$ Which leads to the following formula. $$a_{n} = a_{1}\cdot r^{n - 1}$$ Let's look at some examples.
Example #2: Find a₁₂ and a_n. $$a_{1}=-1, r=3$$ To find a₁₂, let's plug into our formula. $$a_{n}=a_{1}\cdot r^{n - 1}$$ $$a_{12}=-1 \cdot 3^{11} = -177{,}147$$ To find the formula for the general term or a_n, we plug in for a₁ and r. $$a_{n}=-1 \cdot 3^{n - 1}=-3^{n - 1}$$ Example #3: Find a₁₁ and a_n. $$a_3 = 36, a_6 = 972$$ In this case, we are not given a₁ or r, which are both needed for the formula. Let's plug in what we know. $$a_{n}=a_{1}\cdot r^{n - 1}$$ $$a_3 = a_1 \cdot r^{2} ⇒ 36 = a_1 \cdot r^{2}$$ $$a_6 = a_1 \cdot r^{5} ⇒ 972 = a_1 \cdot r^{5}$$ This leads to the following system: $$1) \, a_1 \cdot r^{2} = 36$$ $$2) \, a_1 \cdot r^{5} = 972$$ Let's solve equation #1 for a₁. $$a_1 = \frac{36}{r^2}$$ Now we can plug in for a₁ in equation #2 and solve for r. $$\frac{36}{r^2} \cdot r^{5} = 972$$ $$36r^3 = 972$$ $$r^3 = \frac{972}{36} = 27$$ $$r = \sqrt[3]{27} = 3$$ Since we know that r = 3, we can plug into equation #1 and find a₁. $$a_1 \cdot 3^{2} = 36$$ $$a_1 \cdot 9 = 36$$ $$a_1 = \frac{36}{9} = 4$$ Now we can find a₁₁ and a_n. $$a_{n} = a_{1} \cdot r^{n - 1}$$ $$a_{11} = 4 \cdot 3^{10} = 236{,}196$$ $$a_{n} = 4 \cdot 3^{n - 1}$$

Sum of the First n Terms of a Geometric Sequence

A geometric series is the sum of the terms of a geometric sequence. S_n is used to denote the sum of the first n terms of a geometric sequence. This is known as the nth partial sum and can be found using the following formula. $$S_{n}=\frac{a_{1}(1 - r^n)}{1 - r}, r ≠ 1$$ Obtaining the formula is quite simple. Let's first write our S_n. $$S_n = a_1 + a_1r + a_1r^2 + \cdots + a_1r^{n - 1}$$ Multiply both sides by r: $$rS_n = a_1r + a_1r^2 + a_1r^3 + \cdots + a_1r^{n}$$ Subtract the second equation from the first equation: $$ \begin{array}{l} S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1} \\ rS_n = \hspace{1.8em} a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1} + a_1 r^{n} \\ \hline S_n - rS_n = a_1 - a_1 r^{n} \end{array} $$ Factor out S_n on the left and a₁ on the right. $$S_n(1 - r) = a_1(1 - r^n)$$ To solve for S_n, divide both sides by (1 - r). Since we can't divide by zero, we need to add the restriction that r is not 1. $$S_n = \frac{a_1(1 - r^n)}{1 - r}, r ≠ 1$$ Note: In the special case where r = 1, then every term in the sequence is equal to a₁. $$\text{if} \, r = 1 \, \text{then} \, S_n = n \cdot a_1$$ Let's look at an example.
Example #4: Evaluate each series. $$\sum_{i=1}^{8}2 \cdot (-4)^{i - 1}$$ Here, a₁ = 2 and r = -4. Let's plug into our formula: $$S_{n}=\frac{a_{1}(1 - r^n)}{1 - r}$$ $$S_{8}=\frac{2(1 - (-4)^8)}{1 - (-4)}$$ $$S_{8}=\frac{2(1 - 65{,}536)}{1 + 4}$$ $$S_{8}=\frac{2 \cdot -65{,}535}{5}$$ $$S_{8}=2 \cdot \frac{-65{,}535}{5}$$ $$S_{8}=2 \cdot -13{,}107$$ $$S_{8}=-26{,}214$$

Sum of the Terms of an Infinite Geometric Sequence

$$a_1 + a_1r + a_1r^2 + \cdots + a_1r^{n - 1} + \cdots$$ The above is known as an infinite geometric series. In order to modify our earlier formula, let's consider the following. $$S_n = \frac{a_1(1 - r^n)}{1 - r}, r ≠ 1$$ Suppose that r = 1/3, what happens to the term rⁿ as n gets really large?

Powers of 1/3:

n	(1/3)n	Decimal
1	1/3	0.3
2	1/9	0.1
3	1/27	0.037
4	1/81	≈ 0.0123
5	1/243	≈ 0.0041
6	1/729	≈ 0.0014

Notice that rⁿ is approaching 0 as n increases.
For every positive integer n: $$\text{if} \, {-}1 < r < 1, \, \text{then} \, \lim_{n \to \infty} r^{n} = 0$$ Using this logic, we can plug in a 0 for rⁿ and update our earlier formula. $$\lim_{n \to \infty} S_n = \frac{a_1(1 - 0)}{1 - r} = \frac{a_1}{1 - r}$$ The quotient (a₁)/(1 - r) is called the sum of the terms of an infinite geometric sequence. $$S_{\infty} = \frac{a_1}{1 - r}, -1 < r < 1$$ We could also write this using the summation notation: $$\sum_{i = 1}^{\infty} a_i = \frac{a_1}{1 - r}, -1 < r < 1$$ Note: If |r| ≥ 1, the infinite series does not have a sum. Let's look at an example.
Example #5: Evaluate each series. $$\sum_{i = 1}^{\infty} -3 \cdot \left(-\frac{4}{5}\right)^{i - 1}$$ Here, a₁ = -3 and r = -4/5. Let's plug into our formula. $$S_{\infty} = \frac{-3}{1 - \left(-\frac{4}{5}\right)}$$ $$S_{\infty} = \frac{-3}{1 + \left(\frac{4}{5}\right)}$$ $$S_{\infty} = \frac{-3}{\frac{9}{5}}$$ $$S_{\infty} = -3 \cdot \frac{5}{9}$$ $$S_{\infty} = -\frac{5}{3}$$

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