5.1.6 geometric sequences and sums

Geometric Sequences

A sequence a1, a2 , a3 , … is an geometric sequence if

an = crn, i.e. it is defined by an exponential formula.

Geometric Sequences

Example A. The sequence of powers of 2

a1= 2, a2= 4, a3= 8, a4= 16, …

is an geometric sequence because an = 2n.



Geometric Sequences


a1= 2, a2= 4, a3= 8, a4= 16, …


Fact: If a1, a2 , a3 , …an = c*rn is a geometric sequence, then

the ratio between any two consecutive terms is r.



Geometric Sequences


a1= 2, a2= 4, a3= 8, a4= 16, …




Geometric Sequences

The converse of this fact is also true.




a1= 2, a2= 4, a3= 8, a4= 16, …




Geometric Sequences

The converse of this fact is also true. Below is the formula

that we used for working with geometric sequences.




a1= 2, a2= 4, a3= 8, a4= 16, …




Geometric Sequences



For example, from the sequence above we see that

16/8 = 8/4 = 4/2 = 2 = the ratio r




a1= 2, a2= 4, a3= 8, a4= 16, …




Geometric Sequences

Theorem: If a1, a2 , a3 , …an is a sequence such that

an+1 / an = r for all n, then a1, a2, a3,… is an geometric

sequence and an = a1*rn–1.




16/8 = 8/4 = 4/2 = 2 = the ratio r




a1= 2, a2= 4, a3= 8, a4= 16, …




Geometric Sequences

Theorem: If a1, a2 , a3 , …an is a sequence such that

an+1 / an = r for all n, then a1, a2, a3,… is an geometric

sequence and an = a1*rn–1.

This is the general formula for geometric sequences.




16/8 = 8/4 = 4/2 = 2 = the ratio r



Geometric SequencesGiven the description of a geometric sequence, we use the

general formula to find the specific formula for that sequence.

Example A. The sequence 2, 6, 18, 54, … is an geometric

sequence because 6/2 = 18/6




sequence because 6/2 = 18/6 = 54/18




sequence because 6/2 = 18/6 = 54/18 = … = 3 = r.





Since a1 = 2, set them in the general formula of the

geometric sequences an = a1r n – 1






geometric sequences an = a1r n – 1 , we get the specific

formula for this sequence an = 2(3n – 1).








Geometric Sequences

If a1, a2 , a3 , …an is a geometric sequence such that the

terms alternate between positive and negative signs,

then r is negative.

Given the description of a geometric sequence, we use the







Geometric Sequences



then r is negative.

Example B. The sequence 2/3, –1, 3/2, –9/4, … is a geometric

sequence because

–1/(2/3) = (3/2) / (–1)








Geometric Sequences



then r is negative.


sequence because

–1/(2/3) = (3/2) / (–1) = (–9/4) /(3/2)








Geometric Sequences



then r is negative.


sequence because

–1/(2/3) = (3/2) / (–1) = (–9/4) /(3/2) = … = –3/2 = r.








Geometric Sequences



then r is negative.


sequence because

–1/(2/3) = (3/2) / (–1) = (–9/4) /(3/2) = … = –3/2 = r.

Since a1 = 2/3, the specific formula is

an = ( )n–1 23 2

–3



Geometric SequencesTo use the geometric general formula to find the specific

formula, we need the first term a1 and the ratio r.

Example C. Given that a1, a2 , a3 , …is a geometric sequence

with r = –2 and a5 = 12,




with r = –2 and a5 = 12,

a. find a1




with r = –2 and a5 = 12,

a. find a1

By that the general geometric formula

an = a1r n – 1, we get

a5 = a1(–2)(5 – 1)




with r = –2 and a5 = 12,

a. find a1



a5 = a1(–2)(5 – 1) = 12




with r = –2 and a5 = 12,

a. find a1



a5 = a1(–2)(5 – 1) = 12

a1(–2)4 = 12




with r = –2 and a5 = 12,

a. find a1



a5 = a1(–2)(5 – 1) = 12

a1(–2)4 = 12

16a1 = 12




with r = –2 and a5 = 12,

a. find a1



a5 = a1(–2)(5 – 1) = 12

a1(–2)4 = 12

16a1 = 12

a1 = 12/16 = ¾




with r = –2 and a5 = 12,

a. find a1



a5 = a1(–2)(5 – 1) = 12

a1(–2)4 = 12

16a1 = 12

a1 = 12/16 = ¾



b. find the specific equation.


with r = –2 and a5 = 12,

a. find a1



a5 = a1(–2)(5 – 1) = 12

a1(–2)4 = 12

16a1 = 12

a1 = 12/16 = ¾




Set a1 = ¾ and r = –2 into the general formula an = a1rn – 1 ,


with r = –2 and a5 = 12,

a. find a1



a5 = a1(–2)(5 – 1) = 12

a1(–2)4 = 12

16a1 = 12

a1 = 12/16 = ¾

34

an= (–2)n–1




Set a1 = ¾ and r = –2 into the general formula an = a1rn – 1 ,

we get the specific formula of this sequence

C. Find a9. Geometric Sequences

C. Find a9. Geometric Sequences

34

Since an= (–2)n–1,

set n = 9, we get

C. Find a9.

34

a9= (–2)9–1

Geometric Sequences

34


set n = 9, we get

C. Find a9.

34

a9= (–2)9–1

a9 = (–2)83

4

Geometric Sequences

34


set n = 9, we get

C. Find a9.

34

a9= (–2)9–1

a9 = (–2)8 = (256) = 192 3

4

Geometric Sequences

34


3

4

set n = 9, we get

C. Find a9.

34

a9= (–2)9–1

a9 = (–2)8 = (256) = 192 3

4

Geometric Sequences

34


3

4

Example D. Given that a1, a2 , a3 , …is an geometric

sequence with a3 = –2 and a6 = 54,

a. find r and a1

set n = 9, we get

C. Find a9.

34

a9= (–2)9–1

a9 = (–2)8 = (256) = 192 3

4

Geometric Sequences

34


3

4



a. find r and a1

Given that the general geometric formula an = a1rn – 1,

we have

a3 = –2 = a1r3–1

set n = 9, we get

C. Find a9.

34

a9= (–2)9–1

a9 = (–2)8 = (256) = 192 3

4

Geometric Sequences

34


3

4



a. find r and a1


we have

a3 = –2 = a1r3–1 and a6 = 54 = a1r

6–1

set n = 9, we get

C. Find a9.

34

a9= (–2)9–1

a9 = (–2)8 = (256) = 192 3

4

Geometric Sequences

34


3

4



a. find r and a1


we have

a3 = –2 = a1r3–1 and a6 = 54 = a1r

6–1

–2 = a1r2

set n = 9, we get

C. Find a9.

34

a9= (–2)9–1

a9 = (–2)8 = (256) = 192 3

4

Geometric Sequences

34


3

4



a. find r and a1


we have

a3 = –2 = a1r3–1 and a6 = 54 = a1r

6–1

–2 = a1r2 54 = a1r

5

set n = 9, we get

C. Find a9.

34

a9= (–2)9–1

a9 = (–2)8 = (256) = 192 3

4

Geometric Sequences

34


3

4



a. find r and a1


we have

a3 = –2 = a1r3–1 and a6 = 54 = a1r

6–1

–2 = a1r2 54 = a1r

5

Divide these equations: 54

–2=

a1r5

a1r2

54

–2=

a1r5

a1r2

Geometric Sequences

54

–2=

a1r5

a1r2

–27

Geometric Sequences

54

–2=

a1r5

a1r2

–273 = 5–2

Geometric Sequences

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

Geometric Sequences

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

–3 = r

Geometric Sequences

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

–3 = r

Put r = –3 into the equation –2 = a1r2

Geometric Sequences

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

–3 = r


Hence –2 = a1(–3)2

Geometric Sequences

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

–3 = r


Hence –2 = a1(–3)2

–2 = a19

Geometric Sequences

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

–3 = r


Hence –2 = a1(–3)2

–2 = a19

–2/9 = a1

Geometric Sequences

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

–3 = r


Hence –2 = a1(–3)2

–2 = a19

–2/9 = a1

Geometric Sequences

b. Find the specific formula and a2

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

–3 = r


Hence –2 = a1(–3)2

–2 = a19

–2/9 = a1

Geometric Sequences


Use the general geometric formula an = a1rn – 1,

set a1 = –2/9, and r = –3

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

–3 = r


Hence –2 = a1(–3)2

–2 = a19

–2/9 = a1

Geometric Sequences



set a1 = –2/9, and r = –3 we have the specific formula

–29

an = (–3)n–1

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

–3 = r


Hence –2 = a1(–3)2

–2 = a19

–2/9 = a1

Geometric Sequences




–29

an = (–3)n–1

–29

(–3) 2–1

To find a2, set n = 2, we get

a2 =

54

–2=

a1r5

a1r2

–273 = 5–2

–27 = r3

–3 = r


Hence –2 = a1(–3)2

–2 = a19

–2/9 = a1

Geometric Sequences




–29

an = (–3)n–1

–29

(–3) 2–1

To find a2, set n = 2, we get

–29

a2 = 3

23= (–3) =

Geometric Sequences

Sum of geometric sequences

We observe the algebraic patterns:

(1 – r)(1 + r) = 1 – r2

Geometric Sequences



(1 – r)(1 + r) = 1 – r2

(1 – r)(1 + r + r2) = 1 – r3

Geometric Sequences



(1 – r)(1 + r) = 1 – r2

(1 – r)(1 + r + r2) = 1 – r3

(1 – r)(1 + r + r2 + r3) = 1 – r4

Geometric Sequences



(1 – r)(1 + r) = 1 – r2

(1 – r)(1 + r + r2) = 1 – r3

(1 – r)(1 + r + r2 + r3) = 1 – r4

(1 – r)(1 + r + r2 + r3 + r4) = 1 – r5

Geometric Sequences



(1 – r)(1 + r) = 1 – r2

(1 – r)(1 + r + r2) = 1 – r3

(1 – r)(1 + r + r2 + r3) = 1 – r4

(1 – r)(1 + r + r2 + r3 + r4) = 1 – r5

...

…

(1 – r)(1 + r + r2 … + rn–1) = 1 – rn

Geometric Sequences



(1 – r)(1 + r) = 1 – r2

(1 – r)(1 + r + r2) = 1 – r3

(1 – r)(1 + r + r2 + r3) = 1 – r4

(1 – r)(1 + r + r2 + r3 + r4) = 1 – r5

...

…

(1 – r)(1 + r + r2 … + rn–1) = 1 – rn

Hence 1 + r + r2 + … + rn–1 = 1 – rn

1 – r

Geometric Sequences



(1 – r)(1 + r) = 1 – r2

(1 – r)(1 + r + r2) = 1 – r3

(1 – r)(1 + r + r2 + r3) = 1 – r4

(1 – r)(1 + r + r2 + r3 + r4) = 1 – r5

...

…

(1 – r)(1 + r + r2 … + rn–1) = 1 – rn

Hence 1 + r + r2 + … + rn–1 = 1 – rn

1 – r

Geometric Sequences


Therefore a + ar + ar2 + … +arn–1

n terms


(1 – r)(1 + r) = 1 – r2

(1 – r)(1 + r + r2) = 1 – r3

(1 – r)(1 + r + r2 + r3) = 1 – r4

(1 – r)(1 + r + r2 + r3 + r4) = 1 – r5

...

…

(1 – r)(1 + r + r2 … + rn–1) = 1 – rn

Hence 1 + r + r2 + … + rn–1 = 1 – rn

1 – r

Geometric Sequences



= a (1 + r + r2 + … + r n–1)

n terms


(1 – r)(1 + r) = 1 – r2

(1 – r)(1 + r + r2) = 1 – r3

(1 – r)(1 + r + r2 + r3) = 1 – r4

(1 – r)(1 + r + r2 + r3 + r4) = 1 – r5

...

…

(1 – r)(1 + r + r2 … + rn–1) = 1 – rn

Hence 1 + r + r2 + … + rn–1 = 1 – rn

1 – r

Geometric Sequences


a1 – rn

1 – r


= a (1 + r + r2 + … + r n–1)

n terms

=

Geometric Sequences

= a1 – rn

1 – r

The Sum of the First n Terms of a Geometric Sequence

a + ar + ar2 + … +arn–1

Geometric Sequences

Example E. Find the geometric sum :

2/3 + (–1) + 3/2 + … + (–81/16)

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)

We have a = 2/3 and r = –3/2,

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)

We have a = 2/3 and r = –3/2, and an = –81/16.

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)

We have a = 2/3 and r = –3/2, and an = –81/16. We need the

number of terms.

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)


number of terms. Put a and r in the general formula we get the

specific formula23

– 32

an= ( ) n–1

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)



specific formula23

– 32

an= ( ) n–1

To find n, set an = –8116

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)



specific formula23

– 32

an= ( ) n–1

To find n, set an = =23

– 32

( ) n – 1 –8116

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)



specific formula23

– 32

an= ( ) n–1


– 32

( ) n – 1 –8116

– 32

= ( ) n – 1 –24332

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)



specific formula23

– 32

an= ( ) n–1


– 32

( ) n – 1 –8116

– 32

= ( ) n – 1 –24332

Compare the denominators we see that 32 = 2n – 1.

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)



specific formula23

– 32

an= ( ) n–1


– 32

( ) n – 1 –8116

– 32

= ( ) n – 1 –24332


Since 32 = 25 = 2n – 1

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)



specific formula23

– 32

an= ( ) n–1


– 32

( ) n – 1 –8116

– 32

= ( ) n – 1 –24332


Since 32 = 25 = 2n – 1

n – 1 = 5

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)



specific formula23

– 32

an= ( ) n–1


– 32

( ) n – 1 –8116

– 32

= ( ) n – 1 –24332


Since 32 = 25 = 2n – 1

n – 1 = 5

n = 6

= a1 – rn

1 – r


a + ar + ar2 + … +arn–1

Therefore there are 6 terms in the sum,

2/3 + (–1) + 3/2 + … + (–81/16)

Geometric Sequences


2/3 + (–1) + 3/2 + … + (–81/16)

Geometric Sequences

Set a = 2/3, r = –3/2 and n = 6 in the formula 1 – rn

1 – rS = a


2/3 + (–1) + 3/2 + … + (–81/16)

S = 23

1 – (–3/2)6

1 – (–3/2)

Geometric Sequences


1 – rS = a

we get the sum S


2/3 + (–1) + 3/2 + … + (–81/16)

S = 23

1 – (–3/2)6

1 – (–3/2)

= 23

1 – (729/64)

1 + (3/2)

Geometric Sequences


1 – rS = a

we get the sum S


2/3 + (–1) + 3/2 + … + (–81/16)

S = 23

1 – (–3/2)6

1 – (–3/2)

= 23

1 – (729/64)

1 + (3/2)

=23

–665/645/2

Geometric Sequences


1 – rS = a

we get the sum S


2/3 + (–1) + 3/2 + … + (–81/16)

S = 23

1 – (–3/2)6

1 – (–3/2)

= 23

1 – (729/64)

1 + (3/2)

=23

–665/645/2

–13348

Geometric Sequences


1 – rS = a

we get the sum S

=


2/3 + (–1) + 3/2 + … + (–81/16)

S = 23

1 – (–3/2)6

1 – (–3/2)

= 23

1 – (729/64)

1 + (3/2)

=23

–665/645/2

–13348

Geometric Sequences


1 – rS = a

we get the sum S

=

The “sum” of infinitely many terms of an arithmetic sequences

is always infinite.


2/3 + (–1) + 3/2 + … + (–81/16)

S = 23

1 – (–3/2)6

1 – (–3/2)

= 23

1 – (729/64)

1 + (3/2)

=23

–665/645/2

–13348

Geometric Sequences


1 – rS = a

we get the sum S

=

Adding of infinitely many terms of

an arithmetic sequences is always infinite.

But it’s possible to add infinitely many terms of

a geometric sequence to obtain a finite sum.

Infinite Sums of Geometric SequencesThe following demonstration illustrates how it’s possible to add

infinitely many positive numbers and obtain a finite sum.

Starting with a rope of length 1,



1




take ½ of it,

1




take ½ of it,

take ½ of

what’s left

or 1/4

1




take ½ of it,

take ½ of

what’s left

or 1/4

take ½ of

what’s left

or 1/8

1




take ½ of it,

take ½ of

what’s left

or 1/4

take ½ of

what’s left

or 1/8

take ½ of

what’s left

or 1/16..

1




we see that

take ½ of it,

take ½ of

what’s left

or 1/4

take ½ of

what’s left

or 1/8

take ½ of

what’s left

or 1/16..

1 2

+1 4

+1 8

+1 16

+ .. = 1

1


Infinite Sums of Geometric Sequences

Hence it’s possible to add infinitely many positive numbers

and obtain a finite sum.

The following demonstration illustrates how it’s possible to add


we see that

take ½ of it,

take ½ of

what’s left

or 1/4

take ½ of

what’s left

or 1/8

take ½ of

what’s left

or 1/16..

1 2

+1 4

+1 8

+1 16

+ .. = 1

1




and obtain a finite sum. In fact, the sum of infinitely many terms

from a geometric sequence with ratio | r | < 1 is always finite.



we see that

take ½ of it,

take ½ of

what’s left

or 1/4

take ½ of

what’s left

or 1/8

take ½ of

what’s left

or 1/16..

1 2

+1 4

+1 8

+1 16

+ .. = 1

1




and obtain a finite sum. In fact, the sum of infinitely many terms

from a geometric sequence with ratio | r | < 1 is always finite.

The example above is the sum of the geometric sequence

½, ¼, 1/8, 1/16, … with r = ½ so their sum is finite.



we see that

take ½ of it,

take ½ of

what’s left

or 1/4

take ½ of

what’s left

or 1/8

take ½ of

what’s left

or 1/16..

1 2

+1 4

+1 8

+1 16

+ .. = 1

1

Infinite Sums of Geometric SequencesIf r > 1, then rn grows larger as n gets larger.


For example, the terms of the geometric sequence 2, 4, 8, 16…

(with r = 2) grows larger as the terms go further out.




We write this as an → ∞ as n → ∞.





So clearly in the cases when r > 1 then

a + ar + ar2 + … + arn + … =±∞ just as 2 + 4 + 8 + ... = ∞






a + ar + ar2 + … + arn + … =±∞ just as 2 + 4 + 8 + ... = ∞

But if 0 < r < 1, then rn →0 hence an shrinks to 0 as n gets larger.






a + ar + ar2 + … + arn + … =±∞ just as 2 + 4 + 8 + ... = ∞


For example, the geometric sequence ½, ¼, 1/8, 1/16, …

(with r = ½) shrinks to 0 as the terms go further out.






a + ar + ar2 + … + arn + … =±∞ just as 2 + 4 + 8 + ... = ∞




We write this as an → 0 as n → ∞.






a + ar + ar2 + … + arn + … =±∞ just as 2 + 4 + 8 + ... = ∞





In particular if | r | < 1 then as n → ∞, we have that r n → 0






a + ar + ar2 + … + arn + … =±∞ just as 2 + 4 + 8 + ... = ∞





In particular if | r | < 1 then as n → ∞, we have that r n → 0 and

a1 – rn

1 – ra + ar + ar2 + … +arn–1 + .. =






a + ar + ar2 + … + arn + … =±∞ just as 2 + 4 + 8 + ... = ∞






a1 – rn

1 – ra + ar + ar2 + … +arn–1 + .. = =a

1 – r

0






The Sum of Infinitely–Many Terms of a Geometric Sequence

Given a geometric sequence a, ar, ar2 … with| r | < 1

a + ar + ar2 + … + arn + … =±∞ just as 2 + 4 + 8 + ... = ∞






a1 – rn

1 – ra + ar + ar2 + … +arn–1 + .. = =a

1 – r

0

a rn = a + ar + ar2 + … = a1 – rn=0

∞then

Geometric SequencesExample F. Find the infinite–geometric sum

3/2 + (–1) + 2/3 + … + (–16/81) + 32/243 + ..


3/2 + (–1) + 2/3 + … + (–16/81) + 32/243 + ..

We have a = 3/2 and r = –2/3, hence

3/2 + (–1) + 2/3 + … + (–16/81) + 32/243 + ..

=3/2

1 – (–2/3)


3/2 + (–1) + 2/3 + … + (–16/81) + 32/243 + ..


3/2 + (–1) + 2/3 + … + (–16/81) + 32/243 + ..

=3/2

1 – (–2/3)= (3/2) (3/5) = 9/10

Geometric Sequences

Example G. The area of the

cross–section of a shell may

be approximated with summing

geometric sequences.

Find the cross sectional area of

all the chambers assuming the

cross–sectional area of each

chamber is 1.15 times of the

previous one.google source

15 cm2

Example F. Find the infinite–geometric sum

3/2 + (–1) + 2/3 + … + (–16/81) + 32/243 + ..


3/2 + (–1) + 2/3 + … + (–16/81) + 32/243 + ..

=3/2

1 – (–2/3)= (3/2) (3/5) = 9/10

https://www.google.com/search?q=cross+section+of+shells+geometric+progressions&source=lnms&tbm=isch&sa=X&ei=bJVBVfXQMpPrgwTXzoDwCw&ved=0CAcQ_AUoAQ&biw=1366&bih=657#imgdii=OySWV49N5g_FMM%3A%3BOySWV49N5g_FMM%3A%3BIoyRJYMG2pXrOM%3A&imgrc=OySWV49N5g_FMM%253A%3BALUd8bPrDUWi6M%3Bhttp%253A%252F%252Fwww.spirasolaris.ca%252Fcnaut3c.gif%3Bhttp%253A%252F%252Fwww.spirasolaris.ca%252Fllight.html%3B349%3B433

Geometric SequencesAssuming the ratio of 1.15

is the cross–sectional areas of

the successive chambers,

the areas of the chambers form

a geometric sequence,

starting with the first area of

15 cm2 with r = 1/1.15.

google source

15 cm2








15 cm2 with r = 1/1.15.

google source

15 cm2

15 + 15(1/1.15) + 15(1/1.15)2 + 15(1/1.15)23 + ...

Hence the approximate total

area is the infinite sum:








15 cm2 with r = 1/1.15.

google source

15 cm2

n=0

∞15 + 15(1/1.15) + 15(1/1.15)2 + 15(1/1.15)23 + ...

= 15(1/1.15)n










15 cm2 with r = 1/1.15.

google source

15 cm2



151 – (1/1.15)

n=0

∞15 + 15(1/1.15) + 15(1/1.15)2 + 15(1/1.15)23 + ...

= 15(1/1.15)n

=

= 115 cm2


Geometric SequencesHW.

Given that a1, a2 , a3 , …is a geometric sequence find a1, r,

and the specific formula for the an.

1. a2 = 15, a5 = 405

2. a3 = 3/4, a6 = –2/9

2. a4 = –5/2, a8 = –40

Sum the following geometric sequences.

1. 3 + 6 + 12 + .. + 3072

1. –2 + 6 –18 + .. + 486

1. 6 – 3 + 3/2 – .. + 3/512

5.1.6 geometric sequences and sums

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Transcript of 5.1.6 geometric sequences and sums