1. Chapter 11 Sec 3 Geometric Sequences

2. 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. 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. 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. 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 – 1 108 = a 1 · (3) 3 108 = a 1 · (3) 3 108 = 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. 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 49.92. 3.12, _____, _____, _____, 49.92 a n = a 1 · r n – 1 a n = a 1 · r n – 1 a 5 = 3.12 · r 5 – 1 49.92 = 3.12 r 4 16 = r 4 ±2 = r So… a 1 a 2 a 3 a 4 a 5 6.24 6.24 – 6.24 – 6.24 12.48 24.96 24.96–24.96

7. Chapter 11 Sec 4 Geometric Series

8. 8 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Geometric Series Geometric Sequence Geometric Series. 1, 2, 4, 8, 16 1 + 2 + 4 + 8 + 16 4, –12, 36 4 + (–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. 9 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Example 1 Evaluate

10. 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. 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.

12. Chapter 11 Sec 5 Infinite Geometric Series

13. 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. 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. 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. 16 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Example 2: Sigma Time… Evaluate

17. 17 of 18 Algebra 2 Chapter 11 Sections 3 – 5 Repeeeeating Decimal Write 0.39 as a fraction. S = 0.39 S = 0.393939393939… then 100S = 39.393939393939… Subtract 100S – S 100S = 39.393939393939… Subtract 100S – S – S = 0.393939393939… – S = 0.393939393939… 99S = 39 99S = 39 typo intentional

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