Barisan dan deret 1 bilingual

THE CONCEPT OF SEQUENCE AND SERIES

Hal.: 3 BARISAN DAN DERET AdaptifHal.: 3
Pola Barisan dan Deret Bilangan
Kompetensi Dasar :
Menerapkan konsep barisan dan deret aritmatika
Indikator :
1. Nilai suku ke- n suatu barisan aritmatika ditentukan
menggunakan rumus
2. Jumlah n suku suatu deret aritmatika ditentukan dengan
menggunakan rumus

The Pattern of Sequence and Series Number
Basic Competence:
Applying the concept of arithmetic sequence and
series
Indicator :
1. The value of n-th term in an arithmetic sequence is defined
by formula
2. The sum of n in term of arithmetic sequence is defined by
formula

Saat mengendarai motor, pernahkah kalian mengamati
speedometer pada motor tersebut?
Pada speedometer terdapat angka-angka 0,20, 40, 60, 80, 100,
dan 120 yang menunjukkan kecepatan motor saat kalian
mengendarainya. Angka-angka ini berurutan mulai dari
yang terkecil ke yang terbesar dengan pola tertentu sehingga
membentuk sebuah pola barisan

When you ride a motor cycle, have you ever look at the
speeedometer?
In speedometer,there are numbers of 0,20, 40, 60, 80, 100, and
120 which show the speed of your motor cycle. These numbers
are un order, starts from the smallest to the biggest with certain
pattern, so that it forms a pattern of sequence

Bayangkan anda seorang penumpang taksi. Dia harus membayar biaya buka pintu
Rp 15.000 dan argo Rp 2.500 /km.
15.000 17.500 20.000 22.500 …….
Buka pintu 1 km 2 km 3 km 4 km

Hal.: 8 BARISAN DAN DERET Adaptif
Imagine that you are a taxi passenger. You have to pay the starting fee Rp 15.000
and it charge Rp 2.500 /km.
15.000 17.500 20.000 22.500 …….
Starting fee 1 km 2 km 3 km 4 km

NOTASI SIGMA
Konsep Notasi Sigma
Perhatikan jumlah 6 bilangan ganjil pertama berikut:
1 + 3 + 5 + 7 + 9 + 11 ……….. (1)
Pada bentuk (1)
Suku ke-1 = 1 = 2.1 – 1
Suku ke-2 = 3 = 2.2 – 1
Suku ke-3 = 5 = 2.3 – 1
Suku ke-4 = 7 = 2.4 – 1
Suku ke-5 = 9 = 2.5 – 1
Suku ke-6 = 11 = 2.6 – 1
Secara umum suku ke-k pada (1) dapat dinyatakan
dalam bentuk 2k – 1, k ∈ { 1, 2, 3, 4, 5, 6 }

SIGMA NOTATION
The Concept of Sigma Notation
Look at the sum of the first sixth odd number below:
1 + 3 + 5 + 7 + 9 + 11 ……….. (1)
In the form(1)
The 1st
term = 1 = 2.1 – 1
The 2nd
term= 3 = 2.2 – 1
The 3rd
term = 5 = 2.3 – 1
The 4th
term = 7 = 2.4 – 1
The 5th
term = 9 = 2.5 – 1
The 6th
term = 11 = 2.6 – 1
Generally, the k-th term in (1) can be stated in the form of
2k – 1, k ∈ { 1, 2, 3, 4, 5, 6 }

NOTASI SIGMA
Dengan notasi sigma bentuk penjumlahan (1) dapat
ditulis :
∑=
=+++++
6
1k
1)-(2k1197531

SIGMA NOTATION
In Sigma notation, the addition form (1) can be
written as:
∑=
=+++++
6
1k
1)-(2k1197531

Bentuk ∑
=
−
6
1
)12(
k
k
dibaca “sigma 2k – 1 dari k =1 sampai dengan 6”
atau “jumlah 2k – 1 untuk k = 1 sd k = 6”
1 disebut batas bawah dan
6 disebut batas atas,
k dinamakan indeks
(ada yang menyebut variabel)
∑
=
−−
9
4
)1)3(2(
k
k ∑
=
−
9
4
)72(
k
k
NOTASI SIGMA

In the form
of
∑
=
−
6
1
)12(
k
k
It is read “sigma 2k – 1 from k =1 to 6” or “the sum
of 2k – 1 for k = 1 sd k = 6”
1 is called lower limit and
6 is called upper limit,
k is called index (some people
called it variable)
∑
=
−−
9
4
)1)3(2(
k
k ∑
=
−
9
4
)72(
k
k
SIGMA NOTATION

NOTASI SIGMA
Secara umum

SIGMA NOTATION
Generally

Nyatakan dalam bentuk sigma
1. a + a2
b + a3
b2
+ a4
b3
+ … + a10
b9
∑ −
=
10
1k
)1kbk(a
)142()132()122()112()12(
4
1
+⋅++⋅++⋅++⋅=+∑=k
k
Contoh:
249753 =+++=
Hitung nilai dari:
NOTASI SIGMA

Stated into sigma form
1. a + a2
b + a3
b2
+ a4
b3
+ … + a10
b9
∑ −
=
10
1k
)1kbk(a
)142()132()122()112()12(
4
1
+⋅++⋅++⋅++⋅=+∑=k
k
Example:
249753 =+++=
Define the value of
SIGMA NOTATION

NOTASI SIGMA
nn
n
1n
bCabC...baCbaCbaCa n
1n
33nn
3
22nn
2
1nn
1
n
++++++ −
−
−−−
∑
=
−
n
0r
rrnn
r
baC
2. (a + b)n
=

SIGMA NOTATION
nn
n
1n
bCabC...baCbaCbaCa n
1n
33nn
3
22nn
2
1nn
1
n
++++++ −
−
−−−
∑
=
−
n
0r
rrnn
r
baC
2. (a + b)n
=

Sifat-sifat Notasi Sigma :
, Untuk setiap bilangan bulat a, b dan n
.....1 321
1
n
n
k
aaaaak +++=∑=
∑∑ ==
=
n
mk
n
mk
akCCak.2
∑∑∑ ===
+=+
n
mk
n
mk
n
mk
bkakbkak )(.3
∑∑
+
+==
−
pn
pmk
n
mk
pakak.4
CmnC
n
mk
)1(.5 +−=∑=
∑∑∑ ==
−
=
=+
n
mk
n
pk
p
mk
akakak
1
.6
0.7
1
=∑
=
=
m
mk
ak
NOTASI SIGMA

The properties of sigma notation :
, For every integer a, b and n
.....1 321
1
n
n
k
aaaaak +++=∑=
∑∑ ==
=
n
mk
n
mk
akCCak.2
∑∑∑ ===
+=+
n
mk
n
mk
n
mk
bkakbkak )(.3
∑∑
+
+==
−
pn
pmk
n
mk
pakak.4
CmnC
n
mk
)1(.5 +−=∑=
∑∑∑ ==
−
=
=+
n
mk
n
pk
p
mk
akakak
1
.6
0.7
1
=∑
=
=
m
mk
ak
SIGMA NOTATION

NOTASI SIGMA
Contoh1:
Tunjukkan bahwa
Jawab :
∑∑ ==
+=+
3
1
3
1
)24()24(
jk
ji
30)33.4()22.4()21.4()24(
3
1
=+++++=+∑=i
i
30)23.4()22.4()21.4()24(
3
1
=+++++=+∑=j
j

SIGMA NOTATION
Example 1:
Show that
Answer :
∑∑ ==
+=+
3
1
3
1
)24()24(
jk
ji
30)33.4()22.4()21.4()24(
3
1
=+++++=+∑=i
i
30)23.4()22.4()21.4()24(
3
1
=+++++=+∑=j
j

NOTASI SIGMA
∑∑ ==
+
6
4
2
3
1
2
66
kk
kk
∑∑∑∑ ====
==+
6
1
2
6
1
2
6
4
2
3
1
2
6666
kkkk
kkkk
Hitung nilai dari
Contoh 2 :
Jawab:
= 6 (12
+22
+ 32
+ 42
+ 52
+ 62
)
= 6 (1 + 4 + 9 + 16 + 25 + 36)
= 6.91 = 546

SIGMA NOTATION
∑∑ ==
+
6
4
2
3
1
2
66
kk
kk
∑∑∑∑ ====
==+
6
1
2
6
1
2
6
4
2
3
1
2
6666
kkkk
kkkk
Define the value of
Example 2 :
Answer:
= 6 (12
+22
+ 32
+ 42
+ 52
+ 62
)
= 6 (1 + 4 + 9 + 16 + 25 + 36)
= 6.91 = 546

BARISAN DAN DERET ARITMATIKA
 Bilangan-bilangan berurutan seperti pada speedometer memiliki selisih yang
sama untuk setiap dua suku berurutannya sehingga membentuk suatu barisan
bilangan
 Barisan Aritmatika adalah suatu barisan dengan selisih (beda)
dua suku yang berurutan selalu tetap
Bentuk Umum :
U1, U2, U3, …., Un
a, a + b, a + 2b,…., a + (n-1)b
Pada barisan aritmatika,berlaku Un – Un-1 = b sehingga Un= Un-1 + b

ARITHMETIC SEQUENCE AND SERIES
 The orderly numbers like in speedometer have the same difference for
every two orderly term, so it forms a sequence
 Arithmetic sequence is sequence with difference two orderly term
constant
 The general form is :
U1, U2, U3, …., Un
a, a + b, a + 2b,…., a + (n-1)b
 In arithmetic sequence, we have Un – Un-1 = b, so Un= Un-1 + b

If you start arithmetic sequence with the first term a and difference b, then
you will get this following sequence
The n-th term of arithmetic sequence is Un = a + ( n – 1 )b
Where Un = n-th term
a = the first term
b = difference
n = the term’s quantity
a a + b a + 2b a + 3b …. a + (n-1)b

Hal.: 31 BARISAN DAN DERET AdaptifHl.: 31

If every term of arithmetic sequence is added, then we will get arithmetic
series.
Arithmetic series is the sum of terms of arithmetic sequence
General form :
U1 + U2 + U3 + … + Un atau
a + (a +b) + (a+2b) +… + (a+(n-1)b)
The formula of the sum of the first term in arithmetic series is
Where S = the sum of n-th term
n = the quantity of term
a = the first term
b = difference
= n-th term
( )bna
n
Sn )1(2
2
−+=

Known: the sequence of 5, -2, -9, -16,…., find:
a.The formula of n-th term
b.The 25th
term
Answer:
The difference of two orderly terms in sequence 5,-2, -9,-16 ,…is constant, b= -7,
so that the sequence is an arithmetic sequence
a.The formula of the n-th term in arithmetic sequence is
Un = 5 + ( n – 1 ). -7
Un = 5 + - 7n + 7
Un = -7n + 12
b. The 25th
term of arithmetic sequence is : U12 = - 7.12 + 12
= - 163

Barisan geometri adalah suatu barisan dengan
pembanding (rasio) antara dua suku yang
berurutan selalu tetap.
Ada selembar kertas biru, akan dipotong-potong menjadi dua bagian.
BARISAN DAN DERET GEOMETRI

Geometric sequence is a sequence which has
the constant ratio between two orderly term
There is blue paper. It will cut into two pieces
GEOMETRIC SEQUENCE AND SERIES

Look at the paper part that form a sequence
Every two orderly terms of the sequence have the same ratio
It seems that the ratio of every two orderly terms in the sequence is
always constant. The sequence like this is called geometric
sequence and the comparison of every two orderly term is called
ratio (r)
1 2 4
U1 U2 U3
2....
12
3
1
2
====
−n
n
U
U
U
U
U
U

Geometric sequence is a sequence which have constant
ratio for two orderly term
General form: U1, U2, U3, …., Un atau
a, ar, ar2
, …., arn-1
In geometric sequence
If you start the geometric sequence with the first term a
and the ratio is r, then you get the following sequence
r
U
U
n
n
=
−1
1. −= nn UrsehinggaU

Suku ke-n barisan Geometri adalah :

The n-th term of geometric sequence is :
Start With the
first term a
Multiply with
ratio r
Write the
multiplication
result

Hubungan suku-suku barisan geometri
Seperti dalam barisan Aritmatika hubungan antara suku
yang satu dan suku yang lain dalam barisan geometri dapat
dijelaskan sebagai berikut:
Ambil U12 sebagai contoh :
U12 = a.r11
U12 = a.r9
.r2
= U10. r2
U12 = a.r8
.r3
= U9. r3
U12 = a.r4
.r7
= U5. r7
U12 = a.r3
.r8
= U4.r8
Secara umum dapat dirumuskan bahwa :
Un = Uk. rn-k

The relation of terms in geometric sequence
Like in arithmetic sequence, the relation between terms in
geometric sequence can be explained as follows:
Take U12 as example :
U12 = a.r11
U12 = a.r9
.r2
= U10. r2
U12 = a.r8
.r3
= U9. r3
U12 = a.r4
.r7
= U5. r7
U12 = a.r3
.r8
= U4.r8
Generally, it can be formulated
Un = Uk. rn-k

Geometric series is the sum of terms in geometric sequence
General form
U1 + U2 + U3 + …. + Un
a + ar + ar2
+ ….+ arn-1
The formula of the n sum of the first term in geometric series is
1,
1
)1(
<
−
−
= r
r
ra
S
n
n

Known sequence 27, 9, 3, 1, …..find
a.The formula of the n-th term
b. The 8th
term
Answer:
The ratio of two orderly terms in sequence 27,9,3, 1, …is constant,
so that the sequence is a geometric sequence
a. The formula of the n-th term in geometric sequence is
3
1
=r
1
3
1
27
−






=
n
nU
113
)3.(3 −−
= n
13
3.3 +−
= n
n−
= 4
3

b. The 8th term of geometric sequence is
84
8 3 −
=U
4
3−
=
81
1
=
n
nU −
= 4
3

Deret geometi tak hingga adalah deret geometri yang banyak suku-
sukunya tak hingga.
Jika deret geometri tak hingga dengan -1 < r < 1 , maka jumlah deret
geometri tak hingga tersebut mempunyai limit jumlah (konvergen).
Untuk n = ∞ , rn
mendekati 0
Sehingga S∞ =
Dengan S∞ = Jumlah deret geometri tak hingga
a = Suku pertama
r = rasio
Jika r < -1 atau r > 1 , maka deret geometri tak hingganya akan divergen,
yaitu jumlah suku-sukunya tidak terbatas
Deret Geometri tak hingga
r
a
−1
r
ra
Sn
n
−
−
=
1
)1(

Infinite geometric series is a geometric series which has infinite terms.
If infinite geometric series is -1 < r < 1 , then the sum of geometric series
has sum limit (convergent).
For n = ∞ , rn
is close to 0
So S∞ =
With S∞ = the sum of infinite geometric series
a = the first term
r = ratio
If r < -1 or r > 1 , then the infinite geometric series will be divergent,
means the sum of terms is not limited
Infinite Geometric Series
r
a
−1
r
ra
Sn
n
−
−
=
1
)1(

1. Hitung jumlah deret geometri tak hingga : 18 + 6 + 2 + … . .
Contoh :
3
1
2
3
1
2
===
u
u
u
u
r
27
3
2
18
3
1
1
18
1
==
−
=
−
=∞
r
a
s
Jawab :
a = 18 ;

1. Find the sum of infinite geometric series : 18 + 6 + 2 + … . .
Example :
3
1
2
3
1
2
===
u
u
u
u
r
27
3
2
18
3
1
1
18
1
==
−
=
−
=∞
r
a
s
Answer :
a = 18 ;

2. Sebuah bola elastis dijatuhkan dari ketinggian 2m. Setiap kali memantul dari
lantai, bola mencapai ketinggian ¾ dari ketinggian sebelumnya. Berapakah
panjang lintasan yang dilalui bola hingga berhenti ?
Lihat gambar di samping!
Bola dijatuhkan dari A, maka AB dilalui satu kali,
selanjutnya CD, EF dan seterusnya dilalui dua
kali. Lintasannya membentuk deret geometri
dengan a = 3 dan r = ¾
Panjang lintasan = 2 S∞ - a
2
4
1
2
2
2
4
3
1
2
2
1
2
−












=
−












−
=
−





−
= a
r
a
= 14
Jadi panjang lintasan yang dilalui bola adalah14 m

2. An elastic ball is drop from the height of 2m. Every time it bounce from the
floor, it has ¾ of the previous height. How long is the route that will be passed
by the ball until it stop?
Look at the picture!
The ball is drop from A, so AB is passed only
once. Then CD, EF, etc is passed twice. The
route is in geometric series with a = 3 and r = ¾
the length of the route is= 2 S∞ - a
2
4
1
2
2
2
4
3
1
2
2
1
2
−












=
−












−
=
−





−
= a
r
a
= 14
So, the route length that pass by the ball is 14 m

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