Chapter 11 Sec 3 Geometric Sequences

2 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Geometric Sequence A geometric sequence is a sequence in which each term after the first is found by multiplying the previous term by a constant r called the common ratio.

3 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Example 1 Find the eighth term of a geometric sequence for which a 1 = – 3 and r = – 2. a n = a 1 · r n – 1 a n = a 1 · r n – 1 a 8 = (–3) · (–2) 8 – 1 a 8 = (–3) · (–2) 8 – 1 a 8 = (–3) · (–128) a 8 = (–3) · (–128) a 8 = 384 a 8 = 384

4 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Example 2 Write an equation for the n term of a geometric sequence 3, 12, 48, 192… a n = a 1 · r n – 1 a n = a 1 · r n – 1 a n = (3) · (4) n – 1 a n = (3) · (4) n – 1 So the equations is a n = 3(4) n – 1

5 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Example 3 Find the tenth term of a geometric sequence for which a 4 = 108 and r = 3. a n = a 1 · r n – 1 a 4 = a 1 · (3) 4 – 1 a 4 = a 1 · (3) 4 – = a 1 · (3) = a 1 · (3) = 27a 1 4 = a 1 4 = a 1 a n = a 1 · r n – 1 a 10 = 4 · (3) 10 – 1 a 10 = 4 · (3) 10 – 1 a 10 = 4 · (3) 9 a 10 = 4 · (3) 9 a 10 = 78,732 a 10 = 78,732

6 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Geometric Means As we saw with arithmetic means, you are given two terms of a geometric sequence and are asked to find the terms between, these terms between are called geometric means. Find the three geometric means between 3.12 and , _____, _____, _____, a n = a 1 · r n – 1 a n = a 1 · r n – 1 a 5 = 3.12 · r 5 – = 3.12 r 4 16 = r 4 ±2 = r So… a 1 a 2 a 3 a 4 a – 6.24 – –24.96

Chapter 11 Sec 4 Geometric Series

8 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Geometric Series Geometric Sequence Geometric Series. 1, 2, 4, 8, , –12, (–12) + 36 S n represents the sum of the first n terms of a series. For example, S 4 is the sum of the first four terms.

9 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Example 1 Evaluate

10 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Only have the first and last terms? You can use the formula for finding the nth term in (a n = a 1 · r n – 1 ) conjunction with the sum formula when you don’t know n. when you don’t know n. a n · r = a 1 · r n – 1 · r a n · r = a 1 · r n a n · r = a 1 · r n

11 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Example 3 Find a 1 in a geometric series for which S 8 = 39,360 and r = 3.

Chapter 11 Sec 5 Infinite Geometric Series

13 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Infinite Geometric Series Any geometric series with an infinite number of terms. Consider the infinite geometric series You have already learned to find the sum S n of the first n terms, this is called partial sum for an infinite series. Notice that as n increases, the partial sum levels off and approaches a limit of one. This leveling-off behavior is characteristic of infinite geometric series for which | r | < 1.

14 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Sum of an Infinite Series Lets use the formula for the sum of a finite series to find a formula for an infinite series. If –1 < r < 1, the value if r n will approach 0 as n increases. Therefore the partial sum of the infinite series will approach

15 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Example 1 Find the sum of each infinite geometric series, if it exists. First find the value of r to determine if the sum exists.

16 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Example 2: Sigma Time… Evaluate

17 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Repeeeeating Decimal Write 0.39 as a fraction. S = 0.39 S = … then 100S = … Subtract 100S – S 100S = … Subtract 100S – S – S = … – S = … 99S = 39 99S = 39 typo intentional

18 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Daily Assignment Chapter 11 Sections 3 – 5 Study Guide (SG) Pg 145 – 150 Odd