MATHPOWERTM 12, WESTERN EDITION

6.3 and 6.4

6.3.1

Chapter 6 Sequences and Series

6.3.2

Geometric SequencesA geometric sequence is a sequence where each term isobtained by multiplying the preceding term by a constant, called the common ratio.If tn is a geometric sequence with t1 = a and the commonratio between successive terms, r, then the generalformula is: tn = arn - 1

Where• tn 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 geometricsequence by dividing any term by the previous term:

r tn

tn 1

6.3.3

Geometric Sequences

For the geometric sequence 4, 8, 16, 32, . . .,a) find the general term.b) find the value of t9.

tn = arn - 1

= 4(2)n - 1

= 22(2)n - 1

= 22 + n - 1

tn = 2n + 1

Find the commonratio:

r tn

tn 1

r 84

r 2

tn = 2n + 1

t9 = 29 + 1

t9 = 1024

Use the generalformula:

Use the generalterm:

6.3.4

Geometric Sequences

In a geometric sequence, the sixth term is 972 and the eighth term is 8748. Determine a, r, and tn. t6 = 972972 = ar5

t8 = 87488748 = ar7

t8

t6

ar 7

ar 5

t8

t6

r 2

8748972

r 2

r2 = 9r = ±3

For r = 3: 972 = ar5

972 = a(3)5

972 = 243a 4 = a

For r = -3: 972 = ar5

972 = a(-3)5

972 = -243a -4 = a

tn = arn - 1

tn = 4(3)n - 1

ortn = (-4)(-3)n - 1

a = ±4r = ±3tn = 4(3)n - 1

ortn = (-4)(-3)n - 1

6.3.5

Geometric Sequences - Applications1. 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.

tn = arn - 1

t11 = 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.

tn = arn - 1

t7 = 30 000(0.80)6

= $7864.32

Now 1 2 3 4 5 6 7

The car’s value will be $7864.32.

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, finda) the growth rate. b) the number of students in the first year.

Geometric Sequences - Applications

tn = arn - 1

t7 = 1327(r)6

2036 = 1327 (r)6

1327 4 5 6 7 8 9 10

r 6 20361327

r 20361327

6

r = 1.074

The growth rate is 7.4%.

a)

b) tn = arn - 1

1327 = a(1.074)3 a

13271.0743

a = 1071

There were1071 studentsin the first year.

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 yearswould be $5418.33.

6.4.1

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

10 000(1 0.05)10

P = 6139.13

The initial investment would be $6139.13.

6.4.3

Suggested Questions:Pages 300 and 3011-21 odd, 23 a, 24,27, 28, 30, 34

Pages 304 and 30517-27 odd,28, 30