1. MATHPOWER TM 12, WESTERN EDITION 6.3 and 6.4 6.3.1 Chapter 6 Sequences and Series

2. 6.3.2 Geometric Sequences A geometric sequence is a sequence where each term is obtained by multiplying the preceding term by a constant, called the common ratio. If t n is a geometric sequence with t 1 = a and the common ratio between successive terms, r, then the general formula is: t n = ar n - 1 Where t n is the general term of the geometric sequence, n in the position of the term being considered, a is the first term, and r is the common ratio. You can determine r, the common ratio, for any geometric sequence by dividing any term by the previous term:

3. 6.3.3 Geometric Sequences For the geometric sequence 4, 8, 16, 32,..., a) find the general term. b) find the value of t 9. t n = ar n - 1 = 4(2) n - 1 = 2 2 (2) n - 1 = 2 2 + n - 1 t n = 2 n + 1 Find the common ratio: t n = 2 n + 1 t 9 = 2 9 + 1 t 9 = 1024 Use the general formula: Use the general term:

4. 6.3.4 Geometric Sequences In a geometric sequence, the sixth term is 972 and the eighth term is 8748. Determine a, r, and t n. t 6 = 972 972 = ar 5 t 8 = 8748 8748 = ar 7 r 2 = 9 r = ±3 For r = 3: 972 = ar 5 972 = a(3) 5 972 = 243a 4 = a For r = -3: 972 = ar 5 972 = a(-3) 5 972 = -243a -4 = a t n = ar n - 1 t n = 4(3) n - 1 or t n = (-4)(-3) n - 1 a = ±4 r = ±3 t n = 4(3) n - 1 or t n = (-4)(-3) n - 1

5. 6.3.5 Geometric Sequences - Applications 1. A photocopy machine reduces a picture to 75% of its previous size with each photocopy taken. If it is originally 40 cm long, find its size after the tenth reduction. t n = ar n - 1 t 11 = 40(0.75) 11 - 1 = 2.25 Now 1 2 3 4 5 6 7 8 9 10 11 The picture will be 2.25 cm long. 2. A car that is valued at $30 000 depreciates 20% in value each year. Find its value at the end of six years. t n = ar n - 1 t 7 = 30 000(0.80) 6 = $7864.32 Now 1 2 3 4 5 6 7 The car’s value will be $7864.32.

6. 6.3.6 3. At the end of the fourth year, Archbishop O’Leary High School had a population of 1327 students. At the end of its tenth year, the school had 2036 students. Assuming that the growth rate was consistent, find a)the growth rate. b)the number of students in the first year. Geometric Sequences - Applications t n = ar n - 1 t 7 = 1327(r) 6 2036 = 1327 (r) 6 1327 4 5 6 7 8 9 10 r = 1.074 The growth rate is 7.4%. a) b) t n = ar n - 1 1327 = a(1.074) 3 a = 1071 There were 1071 students in the first year.

7. Compound Interest The formula for compound interest is A = P(1 + i) n. Where: A is the amount of money after investing a principal i is the rate of interest per compounding period n is the number of compounding periods P is the principal (the money invested or borrowed) Example: Find the accumulated amount of $3000 invested at 12% per annum for a period of five years compounded quarterly. A = P(1 + i) n = 3000(1 + 0.03) 20 = 5418.33 A = ? P = 3000 i = 12%/a = 12 ÷ 4 = 3% n = 5 x 4 = 20 The amount after five years would be $5418.33. 6.4.1

8. 6.4.2 Compound Interest What sum invested now will amount to $10 000 in five years at 10%/a compounded semiannually? A = P(1 + i) n 10 000 = P(1 + 0.05) 10 A = 10 000 P = ? i = 10%/a = 10 ÷ 2 = 5% n = 5 x 2 = 10 P = 6139.13 The initial investment would be $6139.13.

9. 6.4.3 Suggested Questions: Pages 300 and 301 1-21 odd, 23 a, 24, 27, 28, 30, 34 Pages 304 and 305 17-27 odd, 28, 30