# barisan geometri

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```					                                   Geometric
Sequences & Series
By: Jeffrey Bivin
Lake Zurich High School

jeff.bivin@lz95.org

Last Updated: October 11, 2005
Geometric Sequences
1, 2, 4, 8, 16, 32, … 2n-1, …

3, 9, 27, 81, 243, … 3n, . . .

n 1
81, 54, 36, 24, 16,        … 81   2 , . .     .
3
34 2 n 1   2 n 1
n 1
 n 5
3           3
Jeff Bivin -- LZHS
nth term
of geometric sequence

an =   a1·r (n-1)

Jeff Bivin -- LZHS
Find the nth term of the
geometric sequence
First term is 2
Common ratio is 3

an =   a1·r(n-1)

an =   2(3)(n-1)

Jeff Bivin -- LZHS
Find the nth term of a
geometric sequence
First term is 128
Common ratio is (1/2)
an = a1·r (n-1)
n 1
1                      1
an  128 
2
an         n 8
2
 1 
an  2  n 1 
7

2 
Jeff Bivin -- LZHS
Find the nth term of the
geometric sequence
First term is 64
Common ratio is (3/2)
an = a1·r(n-1)
n 1
3                  n 1
an  64                3
2          an  n  7
2
3 
n 1
an  2  n1 
6
2 
     
Jeff Bivin -- LZHS
Finding the 10th term
3, 6, 12, 24, 48, . . .
a1 = 3
r=2                 an =  a1·r(n-1)
n = 10                          10-1
an =  3·(2)
an = 3·(2)9

an = 3·(512)
an = 1536
Jeff Bivin -- LZHS
Finding the 8th term
2, -10, 50, -250, 1250, . . .
a1 = 2
r = -5            an =   a1·r(n-1)
n=8                             8-1
an =   2·(-5)
an = 2·(-5)7

an = 2·(-78125)
an = -156250
Jeff Bivin -- LZHS
Sum it up

n
a1  a1r   n
Sn    a r
i 1
1
n 1

1 r

Jeff Bivin -- LZHS
1 + 3 + 9 + 27 + 81 + 243
a1  a1r
n
Sn 
1 r
a1 = 1
1  1 3 6
r=3            Sn 
n=6                    1 3
1  729
Sn 
2
 728
Sn          364
Jeff Bivin -- LZHS
2
4 - 8 + 16 - 32 + 64 – 128 + 256
a1  a1r n
Sn 
1 r
a1 = 4
4  4  ( 2 ) 7
r = -2     Sn 
n=7                1  ( 2 )
4  4(128)
Sn 
3
4  512 516
Sn                     172
Jeff Bivin -- LZHS
3          3
Alternative Sum Formula
a1  a1r     n
Sn 
1 r
n 1
We know that:         an  a1r
n 1
Multiply by r:      an  r  a1r  r
Simplify:             an r  a1r n

a1  an r
Substitute:          Sn 
1 r
Jeff Bivin -- LZHS
Find the sum of the
geometric Series
a1  5                         a1  an r
Sn 
256                         1 r
an                              5 
256  2 
729                                   
729  3 
2                   Sn   
r                                 1
2
3
3
512
5                     10423
Sn         2187
1
Sn 
3
729
Jeff Bivin -- LZHS
10

Evaluate             2
k 1
k
= 2 + 4 + 8+…+1024
a1  a1r
n
Sn 
a1 = 2                        1 r
r=2                        2  22   10
Sn 
n = 10
1 2
an = 1024
2  2  1024
Sn 
1
 2046
Sn            2046
Jeff Bivin -- LZHS
1
8

Evaluate  3  2 j 1 = 3 + 6 + 12 +…+ 384
j 1

a1  a1r
n
Sn 
a1 = 3                    1 r
r=2                    3  3 2  8
Sn 
n=8
1 2
an = 384
3  3  256
Sn 
1
 765
Sn           765
Jeff Bivin -- LZHS
1
Review -- Geometric

nth term        Sum of n terms
a1  a1r   n
Sn 
an =       a1·r (n-1)          1 r
a1  an r
Sn 
1 r

Jeff Bivin -- LZHS
Geometric

Infinite Series

Jeff Bivin -- LZHS
The Magic Flea                 There is
(magnified for easier viewing)   no flea
like a
Magic
Flea

Jeff Bivin -- LZHS
The Magic Flea
(magnified for easier viewing)

...    1 1
32 16
1
8
1
4
1
2

1 1 1  1   1
S           ...  1
2 4 8 16 32
Jeff Bivin -- LZHS
Sum it up -- Infinity


a1
S    a r
i 1
1
n 1

1 r

for        r 1

Jeff Bivin -- LZHS
Remember --The Magic Flea
1 1 1  1    1
S             ...  1
2 4 8 16 32
a1
S 
1 r
a1  2
1             1
S   2

r 21            1 1
2
1
S    2
1
1
2
Jeff Bivin -- LZHS
2 2    2
S  6  2         ...
3 9 27
a1
S 
a1  6           1 r
1
r          S 
6
3           1 1 3

6
S    2
9
3
Jeff Bivin -- LZHS
A Bouncing Ball
rebounds ½ of the distance from which it fell --
What is the total vertical distance that the
ball traveled before coming to rest if it fell
from the top of a 128 feet tall building?

Jeff Bivin -- LZHS
A Bouncing Ball
Downward = 128 + 64 + 32 + 16 + 8 + …
a1   128   128
S             1  256
1 r 1 12   2

Jeff Bivin -- LZHS
A Bouncing Ball
Upward        = 64 + 32 + 16 + 8 + …
a1   64     64
S             1  128
1 r 1 12    2

Jeff Bivin -- LZHS
A Bouncing Ball
Downward = 128 + 64 + 32 + 16 + 8 + … = 256
Upward =       64 + 32 + 16 + 8 + … = 128
TOTAL = 384 ft.

Jeff Bivin -- LZHS
A Bouncing Ball
rebounds 3/5 of the distance from which it fell --
What is the total vertical distance that the ball
traveled before coming to rest if it fell from the top
of a 625 feet tall building?

Jeff Bivin -- LZHS
A Bouncing Ball
Downward = 625 + 375 + 225 + 135 + 81 + …
a1   625   625
S             2  1562 .5
1 r 1 53
5

Jeff Bivin -- LZHS
A Bouncing Ball
Upward       = 375 + 225 + 135 + 81 + …
a1   375   375
S             2  937 .5
1 r 1 53
5

Jeff Bivin -- LZHS
A Bouncing Ball
Downward = 625 + 375 + 225 + 135 + 81 + … = 1562.5
Upward =         375 + 225 + 135 + 81 + … = 937.5

TOTAL = 2500 ft.

Jeff Bivin -- LZHS
Find the sum of the series
S n  .9  .09  .009  .0009  .00009  ...
a1
a1  .9     S 
1 r
r  .1           .9
S 
1  .1
.9
S         1
.9
Jeff Bivin -- LZHS
Fractions - Decimals
.1    1
9           .6    6
9      2
3

.2    2

.7 
9                  7

.3        
3       1          9
9       3

.4    4           .8    8
9
9

.5    5
9
.9    9
9   1

Jeff Bivin -- LZHS
Let’s try again
1
.3 
3
1
.3 
3
1
.3 
3
3
.9      1
3
Jeff Bivin -- LZHS
One more
10 x  9. 9
let       x  .9
subtract    9x  9
x  9 1
9

Jeff Bivin -- LZHS
OK now a series
. 9  .9  .09  .009  .0009  .00009  ...
a1
S 
a1  .9              1 r
r  .1               .9
S 
1  .1

.9
. 9  S     1
.9
Jeff Bivin -- LZHS
.9 = 1
.9 = 1
That’s All
Folks

.9  1
Jeff Bivin -- LZHS

```
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