› jnk15 › math34_s15 › SequenceGeo_slides.pdfGeometric Sequences Math 34: Spring 2015 Geometric Sequences Motivating Examples Review Formula for Geo. Seq. Examples Compound Interest

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review

Formula forGeo. Seq.

Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

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Math 34: Spring 2015

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February 9, 2015

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Geometric Sequences

1 Geometric SequencesMotivating ExamplesReview

2 Formula for Geo. Seq.

3 ExamplesCompound InterestReal World Example

4 Partial SumsFormulaExample

5 Homework

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Motivating Examples

Geometric Sequences will help us answer the following:

An interest-free loan of $12, 000 requires monthlypayments of 15% of the unpaid balance. What is theunpaid or outstanding balance after 18 payments?

Suppose a business makes a $1, 000 profit in its firstmonth and has its monthly profit increase by 10% eachmonth for the next 2 years. How much profit will thebusiness earn in its 24th month? How much profit totalprofit will the business have earned at the end of 2 years?

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Geometric Sequences

A Geometric Sequence is a sequence where the ratiobetween any two consecutive numbers in the sequence is aconstant.

In other words: ak+1/ak = r where r is a constant.

Examples of Geometric Sequences:

1, 4, 16, 64, . . .

has common ratio r = 441 = 4, and 16

4 = 4, and 6416 = 4

32, 16, 8, 4, 2, 1, 12 ,

14 , . . .

has common ratio r = 12

1632 = 1

2 , and 816 = 1

2 , and 48 = 1

2 , etc

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Geometric Sequences




1, 4, 16, 64, . . .

has common ratio r = 441 = 4, and 16

4 = 4, and 6416 = 4

32, 16, 8, 4, 2, 1, 12 ,

14 , . . .


1632 = 1

2 , and 816 = 1

2 , and 48 = 1

2 , etc

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Geometric Sequences




1, 4, 16, 64, . . .


41 = 4, and 16

4 = 4, and 6416 = 4

32, 16, 8, 4, 2, 1, 12 ,

14 , . . .


1632 = 1

2 , and 816 = 1

2 , and 48 = 1

2 , etc

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Geometric Sequences




1, 4, 16, 64, . . . has common ratio r = 441 = 4, and 16

4 = 4, and 6416 = 4

32, 16, 8, 4, 2, 1, 12 ,

14 , . . .


1632 = 1

2 , and 816 = 1

2 , and 48 = 1

2 , etc

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Geometric Sequences





4 = 4, and 6416 = 4

32, 16, 8, 4, 2, 1, 12 ,

14 , . . .


1632 = 1

2 , and 816 = 1

2 , and 48 = 1

2 , etc

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Geometric Sequences





4 = 4, and 6416 = 4

32, 16, 8, 4, 2, 1, 12 ,

14 , . . . has common ratio r = 1

21632 = 1

2 , and 816 = 1

2 , and 48 = 1

2 , etc

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Finding a Formula for a Geometric Sequence

Consider the Geometric Sequence: 1, 4, 16, 64, . . .

Index Sequence(Order) Value

0 11 42 163 62

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




0 1 = 1 = 1 = 1(4)0

1 4 = 1(4) = 1(4) = 1(4)1

2 16 = 4(4) = 1(4)(4) = 1(4)2

3 62 = 16(4) = 1(4)(4)(4) = 1(4)3

So we can write a formula for the (n + 1)st term:an = 1(4)n where the index starts with n = 0

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




0 1 = 1 = 1 = 1(4)0

1 4 = 1(4) = 1(4) = 1(4)1

2 16 = 4(4) = 1(4)(4) = 1(4)2

3 62 = 16(4) = 1(4)(4)(4) = 1(4)3

So we can write a formula for the (n + 1)st term:an = 1(4)n where the index starts with n = 0

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

The Formula for a Geometric Sequence

A geometric sequence can be written as

a0︸︷︷︸a0

, a0(r)︸︷︷︸a1

, a0(r)2︸︷︷︸a2

, . . . , a0(n − 1)︸︷︷︸an−1

The (n + 1)st term of a geometric sequence is(sequence starts with n=0)

an = a0rn

a0 is the first term in the sequence

r is the common ratio (r = a1a0

= a2a1

= . . . )

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Geometric Sequence Other Info

A geometric sequence can be written as

a0︸︷︷︸a0

, a0(r)︸︷︷︸a1

, a0(r)2︸︷︷︸a2

, . . . , a0(n − 1)︸︷︷︸an−1

Note that we start the index with 0, so...

The first term is a0,The second term is a1,The third term is a2,etc.

Each term is r times the previous term:

ak = ak−1 · r

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Geometric Sequence Example:

For the Geometric Sequence 9, 12, 16, 2113 , . . .

1 What is the common ratio?

The common ratio is r = 43

129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,

2 What is the first term in the sequence?

a0 = 9

3 What is the fifth term in the sequence?

a4 = 28.4We can find the next term by multiplying 21.3 by 4

3

4 Find a formula for an. (the n + 1st term in the sequence)

an = 9(43)n

We know Sn = a0(r)n

We know a0 = 9 and r = 43

5 What is the 9th term in the sequence? (Round to 4 decimal places)

9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework





129 = 4

3 ,

and 1612 = 4

3 , and21 1

316 = 4

3 ,


a0 = 9



3


an = 9(43)n

We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework





129 = 4

3 , and 1612 = 4

3 ,

and21 1

316 = 4

3 ,


a0 = 9



3


an = 9(43)n

We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework





129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,


a0 = 9



3


an = 9(43)n

We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework



1 What is the common ratio? The common ratio is r = 43

129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,


a0 = 9



3


an = 9(43)n

We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,

2 What is the first term in the sequence? a0 = 9



3


an = 9(43)n

We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,



a4 = 28.4

We can find the next term by multiplying 21.3 by 43


an = 9(43)n

We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,


3 What is the fifth term in the sequence? a4 = 28.4We can find the next term by multiplying 21.3 by 4

3


an = 9(43)n

We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,



3


an = 9(43)n

We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,



3


an = 9(43)n

We know Sn = a0(r)n

We know a0 = 9

and r = 43


9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,



3


an = 9(43)n

We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,



3

4 Find a formula for an. (the n + 1st term in the sequence) an = 9(43)n

We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,



3


We know Sn = a0(r)n



9th term is a8.

a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,



3


We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,



3


We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19.

a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




129 = 4

3 , and 1612 = 4

3 , and21 1

316 = 4

3 ,



3


We know Sn = a0(r)n



9th term is a8. a8 = 9(43)8 = 89.8985


20th term is a19. a19 = 9(43)19 = 2128.5238

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Geometric Sequences and Compound Interest

Recall that the compound interest formula is:

FV = PV (1 + i)n

Geometric Interest Formula:

an = a0( r )n

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework



FV = PV (1 + i)n


an = a0( r )n

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework



FV = PV (1 + i)n


an = a0( r )n

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework



FV = PV (1 + i)n


an = a0( r )n

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework



FV = PV (1 + i)n


an = a0( r )n

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework



FV = PV (1 + i)n


an = a0( r )n

Compound Interest is a Geometric Sequence:The first term: a0 = PVThe common ratio: r = (1 + i)

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework



FV = PV (1 + i)n


an = a0( r )n

Compound Interest is a Geometric Sequence:The first term: a0 = PVThe common ratio: r = (1 + i)

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Revisiting Our First Compound Interest Example:(From 3.1 Notes)

Suppose Olaf invests $5, 000 in an investment that pays6% interest compounded annually. How much does hehave at the end of each of the first 5 years?

Year Interest earned that year Balance at end of year

1 $5000 ∗ 0.06 ∗ 1 = $300 $5300.00

2 $5300 ∗ .06 ∗ 1 = $318 $5618.00

3 $5618 ∗ .06 ∗ 1 = $337.08 $5955.08

4 $5955.08 ∗ .06 ∗ 1 = $357.30 $6312.38

5 $6312.38 ∗ .06 ∗ 1 = $378.74 $6691.12

So 5000, 5300, 5618, 5955.08, 6312.38, 6691.12, . . .is a geometric sequence,

with a0 = $5300

and r = 1.06 r =$5300

$5000= 1.06

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




1 $5000 ∗ 0.06 ∗ 1 = $300 $5300.00

2 $5300 ∗ .06 ∗ 1 = $318 $5618.00

3 $5618 ∗ .06 ∗ 1 = $337.08 $5955.08

4 $5955.08 ∗ .06 ∗ 1 = $357.30 $6312.38

5 $6312.38 ∗ .06 ∗ 1 = $378.74 $6691.12


with a0 = $5300

and r = 1.06 r =$5300

$5000= 1.06

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




1 $5000 ∗ 0.06 ∗ 1 = $300 $5300.00

2 $5300 ∗ .06 ∗ 1 = $318 $5618.00

3 $5618 ∗ .06 ∗ 1 = $337.08 $5955.08

4 $5955.08 ∗ .06 ∗ 1 = $357.30 $6312.38

5 $6312.38 ∗ .06 ∗ 1 = $378.74 $6691.12


with a0 = $5300

and r = 1.06 r =$5300

$5000= 1.06

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




1 $5000 ∗ 0.06 ∗ 1 = $300 $5300.00

2 $5300 ∗ .06 ∗ 1 = $318 $5618.00

3 $5618 ∗ .06 ∗ 1 = $337.08 $5955.08

4 $5955.08 ∗ .06 ∗ 1 = $357.30 $6312.38

5 $6312.38 ∗ .06 ∗ 1 = $378.74 $6691.12


with a0 = $5300

and r = 1.06 r =$5300

$5000= 1.06

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Real World Examples 1

An interest-free loan of $12, 000 requires monthly payments of15% of the unpaid balance. What is the unpaid or outstandingbalance after 18 payments?

Let’s Work the Remaining Balance for a few months:

Your starting balance (after 0 months) is 12, 000After 1 month, you owe a payment:Your first payment is 15% of $12, 0000.15 ∗ 12000 = 1800Remaining Balance (after 1 month):12, 000 − 1800 = 10, 200Another way to think of this, if you paid of 15% of 12000,you have 85% of 12000 remainingYour second payment is 15% of $10, 2000.15 ∗ 10, 200 = 1530.Remaining balance (after 2 months):10200 − 1530 = 8670

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




Your starting balance (after 0 months) is

12, 000After 1 month, you owe a payment:Your first payment is 15% of $12, 0000.15 ∗ 12000 = 1800Remaining Balance (after 1 month):12, 000 − 1800 = 10, 200Another way to think of this, if you paid of 15% of 12000,you have 85% of 12000 remainingYour second payment is 15% of $10, 2000.15 ∗ 10, 200 = 1530.Remaining balance (after 2 months):10200 − 1530 = 8670

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




Your starting balance (after 0 months) is 12, 000

After 1 month, you owe a payment:Your first payment is 15% of $12, 0000.15 ∗ 12000 = 1800Remaining Balance (after 1 month):12, 000 − 1800 = 10, 200Another way to think of this, if you paid of 15% of 12000,you have 85% of 12000 remainingYour second payment is 15% of $10, 2000.15 ∗ 10, 200 = 1530.Remaining balance (after 2 months):10200 − 1530 = 8670

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




Your starting balance (after 0 months) is 12, 000After 1 month, you owe a payment:

Your first payment is 15% of $12, 0000.15 ∗ 12000 = 1800Remaining Balance (after 1 month):12, 000 − 1800 = 10, 200Another way to think of this, if you paid of 15% of 12000,you have 85% of 12000 remainingYour second payment is 15% of $10, 2000.15 ∗ 10, 200 = 1530.Remaining balance (after 2 months):10200 − 1530 = 8670

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




Your starting balance (after 0 months) is 12, 000After 1 month, you owe a payment:Your first payment is 15% of $12, 0000.15 ∗ 12000 = 1800

Remaining Balance (after 1 month):12, 000 − 1800 = 10, 200Another way to think of this, if you paid of 15% of 12000,you have 85% of 12000 remainingYour second payment is 15% of $10, 2000.15 ∗ 10, 200 = 1530.Remaining balance (after 2 months):10200 − 1530 = 8670

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




Your starting balance (after 0 months) is 12, 000After 1 month, you owe a payment:Your first payment is 15% of $12, 0000.15 ∗ 12000 = 1800Remaining Balance (after 1 month):12, 000 − 1800 = 10, 200

Another way to think of this, if you paid of 15% of 12000,you have 85% of 12000 remainingYour second payment is 15% of $10, 2000.15 ∗ 10, 200 = 1530.Remaining balance (after 2 months):10200 − 1530 = 8670

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




Your starting balance (after 0 months) is 12, 000After 1 month, you owe a payment:Your first payment is 15% of $12, 0000.15 ∗ 12000 = 1800Remaining Balance (after 1 month):12, 000 − 1800 = 10, 200Another way to think of this, if you paid of 15% of 12000,you have 85% of 12000 remaining

Your second payment is 15% of $10, 2000.15 ∗ 10, 200 = 1530.Remaining balance (after 2 months):10200 − 1530 = 8670

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




Your starting balance (after 0 months) is 12, 000After 1 month, you owe a payment:Your first payment is 15% of $12, 0000.15 ∗ 12000 = 1800Remaining Balance (after 1 month):12, 000 − 1800 = 10, 200Another way to think of this, if you paid of 15% of 12000,you have 85% of 12000 remainingYour second payment is 15% of $10, 200

0.15 ∗ 10, 200 = 1530.Remaining balance (after 2 months):10200 − 1530 = 8670

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




Your starting balance (after 0 months) is 12, 000After 1 month, you owe a payment:Your first payment is 15% of $12, 0000.15 ∗ 12000 = 1800Remaining Balance (after 1 month):12, 000 − 1800 = 10, 200Another way to think of this, if you paid of 15% of 12000,you have 85% of 12000 remainingYour second payment is 15% of $10, 2000.15 ∗ 10, 200 = 1530.Remaining balance (after 2 months):10200 − 1530 = 8670

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(after)Month(s) Unpaid Balance

0 $12, 0001 $10, 2002 $8, 670

a1a0

=10, 200

12, 000= 0.85

a2a1

=8670

10200= 0.85

So this is a geometric sequence:

a0 =

12000

r =

0.85

So an =

12, 000(0.85)n

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0 $12, 0001 $10, 2002 $8, 670

a1a0

=10, 200

12, 000= 0.85

a2a1

=8670

10200= 0.85


a0 =

12000

r =

0.85

So an =

12, 000(0.85)n

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0 $12, 0001 $10, 2002 $8, 670

a1a0

=10, 200

12, 000= 0.85

a2a1

=8670

10200= 0.85


a0 =

12000

r =

0.85

So an =

12, 000(0.85)n

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0 $12, 0001 $10, 2002 $8, 670

a1a0

=10, 200

12, 000= 0.85

a2a1

=8670

10200= 0.85


a0 =

12000

r =

0.85

So an =

12, 000(0.85)n

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0 $12, 0001 $10, 2002 $8, 670

a1a0

=10, 200

12, 000= 0.85

a2a1

=8670

10200= 0.85


a0 =

12000

r =

0.85

So an =

12, 000(0.85)n

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0 $12, 0001 $10, 2002 $8, 670

a1a0

=10, 200

12, 000= 0.85

a2a1

=8670

10200= 0.85


a0 = 12000r =

0.85

So an =

12, 000(0.85)n

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0 $12, 0001 $10, 2002 $8, 670

a1a0

=10, 200

12, 000= 0.85

a2a1

=8670

10200= 0.85


a0 = 12000r = 0.85

So an =

12, 000(0.85)n

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0 $12, 0001 $10, 2002 $8, 670

a1a0

=10, 200

12, 000= 0.85

a2a1

=8670

10200= 0.85


a0 = 12000r = 0.85

So an =

12, 000(0.85)n

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0 $12, 0001 $10, 2002 $8, 670

a1a0

=10, 200

12, 000= 0.85

a2a1

=8670

10200= 0.85


a0 = 12000r = 0.85

So an = 12, 000(0.85)n

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So an = 12, 000(0.85)n

Since a1 is balance remaining after month 1, and a2 isbalance remaining after month 2 ....

Balance remaining after 18 months is....

a18 = 12000(0.85)18 = 643.76

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So an = 12, 000(0.85)n


Balance remaining after 18 months is....a18 =

12000(0.85)18 = 643.76

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So an = 12, 000(0.85)n


Balance remaining after 18 months is....a18 = 12000(0.85)18 = 643.76

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Finding Formula for the Sum of the First n Termsof a Geometric Sequence:

Find the nth partial sum of the geometric series

a0︸︷︷︸a0

, a0(r)︸︷︷︸a1

, a0(r)2︸︷︷︸a2

, . . . a0(n − 1)︸︷︷︸an−1

Sn = a0 +a1 +a2 + . . . +an−1

Sn = a0 +a0r +a0(r)2 + . . . +a0rn−1

Now for a clever trickSn = a0 +a0r +a0(r)2 + . . . +a0r

n−1

r(Sn) = r(a0 +a0r +a0(r)2 + . . . +a0r

n−1)

r(Sn) = a0r +a0rr +a0(r)2r + . . . +a0rn−1r

r(Sn) = a0r +a0(r)2 +a0(r)3 + . . . +a0rn

Subtract the 2 Equations....

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a0︸︷︷︸a0

, a0(r)︸︷︷︸a1

, a0(r)2︸︷︷︸a2

, . . . a0(n − 1)︸︷︷︸an−1

Sn = a0 +a1 +a2 + . . . +an−1

Sn = a0 +a0r +a0(r)2 + . . . +a0rn−1


n−1

r(Sn) = r(a0 +a0r +a0(r)2 + . . . +a0r

n−1)


r(Sn) = a0r +a0(r)2 +a0(r)3 + . . . +a0rn


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a0︸︷︷︸a0

, a0(r)︸︷︷︸a1

, a0(r)2︸︷︷︸a2

, . . . a0(n − 1)︸︷︷︸an−1

Sn = a0 +a1 +a2 + . . . +an−1

Sn = a0 +a0r +a0(r)2 + . . . +a0rn−1


n−1

r(Sn) = r(a0 +a0r +a0(r)2 + . . . +a0r

n−1)


r(Sn) = a0r +a0(r)2 +a0(r)3 + . . . +a0rn


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a0︸︷︷︸a0

, a0(r)︸︷︷︸a1

, a0(r)2︸︷︷︸a2

, . . . a0(n − 1)︸︷︷︸an−1

Sn = a0 +a1 +a2 + . . . +an−1

Sn = a0 +a0r +a0(r)2 + . . . +a0rn−1


n−1

r(Sn) = r(a0 +a0r +a0(r)2 + . . . +a0r

n−1)


r(Sn) = a0r +a0(r)2 +a0(r)3 + . . . +a0rn


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Subtract the 2 Equations....Sn = a0 +a0r +a0(r)2 + . . . +a0r

n−1

−r(Sn) = −(a0r +a0(r)2 +a0(r)3 + . . . +a0r

n)

Sn − rSn = a0 −a0rn

Now a bit of Algebra

Sn − rSn = a0 − a0rn

Sn(1 − r) = a0(1 − rn)

Sn(1 − r)

(1 − r)=

a0(1 − rn)

1 − r

Sn =a0(1− rn)

1− r

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n−1

−r(Sn) = −(a0r +a0(r)2 +a0(r)3 + . . . +a0r

n)




Sn(1 − r) = a0(1 − rn)

Sn(1 − r)

(1 − r)=

a0(1 − rn)

1 − r

Sn =a0(1− rn)

1− r

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n−1

−r(Sn) = −(a0r +a0(r)2 +a0(r)3 + . . . +a0r

n)




Sn(1 − r) = a0(1 − rn)

Sn(1 − r)

(1 − r)=

a0(1 − rn)

1 − r

Sn =a0(1− rn)

1− r

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n−1

−r(Sn) = −(a0r +a0(r)2 +a0(r)3 + . . . +a0r

n)




Sn(1 − r) = a0(1 − rn)

Sn(1 − r)

(1 − r)=

a0(1 − rn)

1 − r

Sn =a0(1− rn)

1− r

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Formula for Partial Sum of Geometric Sequence

The sum of the first n terms of a geometric sequence withfirst term a0 and common ratio r is

Sn =a0(1 − rn)

1 − r

As long as r is not equal to 1.

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Partial Sum of Geometric Sequence Example:

Find the 20th partial sum of the Geometric Sequence1000, 1200, 1440, 1728, . . .

Identify a0

= 1000

Identify r

= 1.212001000 = 1.2 1440

1200 = 1.2

Plug into Sn formula.

Sn = a0(1−rn)1−r

Sn = 1000(1−(1.2)n)1−(1.2)

S20 = 1000(1−(1.2)20)1−(1.2)

S20 = $186687.999622 ≈ $186, 688.00

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Identify a0

= 1000

Identify r

= 1.212001000 = 1.2 1440

1200 = 1.2


Sn = a0(1−rn)1−r

Sn = 1000(1−(1.2)n)1−(1.2)

S20 = 1000(1−(1.2)20)1−(1.2)

S20 = $186687.999622 ≈ $186, 688.00

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Homework



Identify a0 = 1000

Identify r

= 1.212001000 = 1.2 1440

1200 = 1.2


Sn = a0(1−rn)1−r

Sn = 1000(1−(1.2)n)1−(1.2)

S20 = 1000(1−(1.2)20)1−(1.2)

S20 = $186687.999622 ≈ $186, 688.00

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Identify a0 = 1000

Identify r

= 1.2

12001000 = 1.2

14401200 = 1.2


Sn = a0(1−rn)1−r

Sn = 1000(1−(1.2)n)1−(1.2)

S20 = 1000(1−(1.2)20)1−(1.2)

S20 = $186687.999622 ≈ $186, 688.00

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Homework



Identify a0 = 1000

Identify r

= 1.2

12001000 = 1.2 1440

1200 = 1.2


Sn = a0(1−rn)1−r

Sn = 1000(1−(1.2)n)1−(1.2)

S20 = 1000(1−(1.2)20)1−(1.2)

S20 = $186687.999622 ≈ $186, 688.00

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Identify a0 = 1000

Identify r = 1.212001000 = 1.2 1440

1200 = 1.2


Sn = a0(1−rn)1−r

Sn = 1000(1−(1.2)n)1−(1.2)

S20 = 1000(1−(1.2)20)1−(1.2)

S20 = $186687.999622 ≈ $186, 688.00

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Identify a0 = 1000

Identify r = 1.212001000 = 1.2 1440

1200 = 1.2


Sn = a0(1−rn)1−r

Sn = 1000(1−(1.2)n)1−(1.2)

S20 = 1000(1−(1.2)20)1−(1.2)

S20 = $186687.999622 ≈ $186, 688.00

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Identify a0 = 1000

Identify r = 1.212001000 = 1.2 1440

1200 = 1.2


Sn = a0(1−rn)1−r

Sn = 1000(1−(1.2)n)1−(1.2)

S20 = 1000(1−(1.2)20)1−(1.2)

S20 = $186687.999622 ≈ $186, 688.00

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Identify a0 = 1000

Identify r = 1.212001000 = 1.2 1440

1200 = 1.2


Sn = a0(1−rn)1−r

Sn = 1000(1−(1.2)n)1−(1.2)

S20 = 1000(1−(1.2)20)1−(1.2)

S20 = $186687.999622 ≈ $186, 688.00

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Identify a0 = 1000

Identify r = 1.212001000 = 1.2 1440

1200 = 1.2


Sn = a0(1−rn)1−r

Sn = 1000(1−(1.2)n)1−(1.2)

S20 = 1000(1−(1.2)20)1−(1.2)

S20 = $186687.999622 ≈ $186, 688.00

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Real World Examples 2:

Suppose a business makes a $1, 000 profit in its first monthand has its monthly profit increase by 10% each month for thenext 2 years. How much profit will the business earn in its 24th

month? How much profit total profit will the business haveearned at the end of 2 years?

Work out a few months: Remember, we need the index tostart at 0.

Month Index Monthly Profit

Month 1 0 $1, 000Month 2 1Month 3 2

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Month 1 0 $1, 000Month 2 1 $1, 100Month 3 2 $1, 210

Realize this is a Geometric Series

Identify a0

= 1000

Identify r

= 1.1

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Homework








Identify a0

= 1000

Identify r

= 1.1

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Identify a0

= 1000

Identify r

= 1.1

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Homework








Identify a0 = 1000Identify r

= 1.1

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Identify a0 = 1000Identify r = 1.1

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Suppose a business makes a $1, 000 profit in its first month and has itsmonthly profit increase by 10% each month for the next 2 years. Howmuch profit will the business earn in its 24th month? How much profittotal profit will the business have earned at the end of 2 years?

a0 = 1000 and r = 1.1

How much profit will the business earn in its 24th month?

We want one of the terns of the geometric sequenceThis will be answered by an anSince the index starts at n = 0, the 24th term is a23a23 = 1000(1.1)23 = 8, 954.30

How much profit total profit will the business have earnedat the end of 2 years?

This will be answered by a partial sumWe will answer it with S24S24 = 1000(1−1.124)

(1−1.1) = $88, 497.33

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a0 = 1000 and r = 1.1

How much profit will the business earn in its 24th month?We want one of the terns of the geometric sequenceThis will be answered by an an

Since the index starts at n = 0, the 24th term is a23a23 = 1000(1.1)23 = 8, 954.30



(1−1.1) = $88, 497.33

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a0 = 1000 and r = 1.1

How much profit will the business earn in its 24th month?We want one of the terns of the geometric sequenceThis will be answered by an anSince the index starts at n = 0, the 24th term is a23

a23 = 1000(1.1)23 = 8, 954.30



(1−1.1) = $88, 497.33

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a0 = 1000 and r = 1.1

How much profit will the business earn in its 24th month?We want one of the terns of the geometric sequenceThis will be answered by an anSince the index starts at n = 0, the 24th term is a23a23 = 1000(1.1)23 = 8, 954.30



(1−1.1) = $88, 497.33

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a0 = 1000 and r = 1.1


How much profit total profit will the business have earnedat the end of 2 years?This will be answered by a partial sum

We will answer it with S24S24 = 1000(1−1.124)

(1−1.1) = $88, 497.33

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a0 = 1000 and r = 1.1


How much profit total profit will the business have earnedat the end of 2 years?This will be answered by a partial sumWe will answer it with S24

S24 = 1000(1−1.124)(1−1.1) = $88, 497.33

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a0 = 1000 and r = 1.1


How much profit total profit will the business have earnedat the end of 2 years?This will be answered by a partial sumWe will answer it with S24S24 = 1000(1−1.124)

(1−1.1) = $88, 497.33

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Shortcuts to Finding r

Sometimes there is an easier way to find r rather thanworking out several terms and checking the ratio.

If each term in the sequence is a certain percent morethan the previous term:

r = 1 + p

(where p is the percent, converted to a decimal)

If each term in the sequence is a certain percent less thanthe previous term:

r = 1 − p

(where p is the percent, converted to a decimal)

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You start advertising your dog grooming business on a new social network

called Woofer. Your advertising cost in January $100. Since the social

network is growing in popularity, your advertising cost in February are 8%

higher. You assume this pattern will continue, and each month your

advertising cost will be 8% higher than the previous month.

1 What are your advertising costs in December?

This will be a Geo Seriesa0=advert. costs in Jan, a2 = advert. costs in Feb...a11=advertising costs in Deca0 = 100 and r = 1.08a11 = 100(1.08)11 = 233.16

2 How much will you spend in advertising over the year?

This is a partial sum, we want a0 + a1 + · · · + a11S12 will be the answer

S12 = 100(1−1.0812)(1−1.08) = 1, 897.71

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1 What are your advertising costs in December?This will be a Geo Series

a0=advert. costs in Jan, a2 = advert. costs in Feb...a11=advertising costs in Deca0 = 100 and r = 1.08a11 = 100(1.08)11 = 233.16



S12 = 100(1−1.0812)(1−1.08) = 1, 897.71

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1 What are your advertising costs in December?This will be a Geo Seriesa0=advert. costs in Jan, a2 = advert. costs in Feb...a11=advertising costs in Dec

a0 = 100 and r = 1.08a11 = 100(1.08)11 = 233.16



S12 = 100(1−1.0812)(1−1.08) = 1, 897.71

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Example

Homework







1 What are your advertising costs in December?This will be a Geo Seriesa0=advert. costs in Jan, a2 = advert. costs in Feb...a11=advertising costs in Deca0 = 100 and r = 1.08

a11 = 100(1.08)11 = 233.16



S12 = 100(1−1.0812)(1−1.08) = 1, 897.71

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework







1 What are your advertising costs in December?This will be a Geo Seriesa0=advert. costs in Jan, a2 = advert. costs in Feb...a11=advertising costs in Deca0 = 100 and r = 1.08a11 = 100(1.08)11 = 233.16



S12 = 100(1−1.0812)(1−1.08) = 1, 897.71

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework








2 How much will you spend in advertising over the year?This is a partial sum, we want a0 + a1 + · · · + a11

S12 will be the answer

S12 = 100(1−1.0812)(1−1.08) = 1, 897.71

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework








2 How much will you spend in advertising over the year?This is a partial sum, we want a0 + a1 + · · · + a11S12 will be the answer

S12 = 100(1−1.0812)(1−1.08) = 1, 897.71

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework








2 How much will you spend in advertising over the year?This is a partial sum, we want a0 + a1 + · · · + a11S12 will be the answer

S12 = 100(1−1.0812)(1−1.08) = 1, 897.71

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework


A business has a profit of $25, 000 in the first year and thenloses 5% each year for the next seven years.

This will be a Geo Series. a0 = 25, 000 and r = 0.95

1 What is the business’ profit in the 4th year?

We’ll answer with a term of the sequenceSince the profit in the first year is a0, the profit in the 4thyear is a3a3 = 25000(0.95)3 = $21434.38

2 What is the total profit after 7 years?

Since this is about adding up the profit in each of the first7 years...We answer with S7S7 = 25000(1−0.957)

(1−0.95) = $150, 831.35

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework


A business has a profit of $25, 000 in the first year and thenloses 5% each year for the next seven years.This will be a Geo Series. a0 = 25, 000 and r = 0.95

1 What is the business’ profit in the 4th year?

We’ll answer with a term of the sequenceSince the profit in the first year is a0, the profit in the 4thyear is a3a3 = 25000(0.95)3 = $21434.38



(1−0.95) = $150, 831.35

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework



1 What is the business’ profit in the 4th year?We’ll answer with a term of the sequenceSince the profit in the first year is a0, the profit in the 4thyear is a3

a3 = 25000(0.95)3 = $21434.38



(1−0.95) = $150, 831.35

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework



1 What is the business’ profit in the 4th year?We’ll answer with a term of the sequenceSince the profit in the first year is a0, the profit in the 4thyear is a3a3 = 25000(0.95)3 = $21434.38



(1−0.95) = $150, 831.35

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




2 What is the total profit after 7 years?Since this is about adding up the profit in each of the first7 years...

We answer with S7S7 = 25000(1−0.957)

(1−0.95) = $150, 831.35

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




2 What is the total profit after 7 years?Since this is about adding up the profit in each of the first7 years...We answer with S7

S7 = 25000(1−0.957)(1−0.95) = $150, 831.35

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework




2 What is the total profit after 7 years?Since this is about adding up the profit in each of the first7 years...We answer with S7S7 = 25000(1−0.957)

(1−0.95) = $150, 831.35

GeometricSequences

Math 34:Spring 2015

GeometricSequences

MotivatingExamples

Review


Examples

CompoundInterest

Real WorldExample

Partial Sums

Formula

Example

Homework

Homework

It is NOT in your book.

It IS at the end of the printout on the course website.

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› jnk15 › math34_s15 › SequenceGeo_slides.pdfGeometric Sequences Math 34: Spring 2015 Geometric Sequences Motivating Examples Review Formula for Geo. Seq. Examples Compound Interest