jANJANG GEOMETRI
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Transcript of jANJANG GEOMETRI
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JANJANG GEOMETRI
DISEDIAAN OLEH :SITI SHUHADA BINTI SAUFI
NURUL NAJIHAH BINTI ABDUL RAZAK
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SEBUTAN JANJANG GEOMETRI
Dalam Janjang Geometri nisbah untuk mana-mana sebutan (kecuali yang pertama) dengan sebutan yang sebelumnya adalah tetap, nisbah ini disebut nisbah sepunya dengan simbol (r).
• Siri Janjang Geometri T1, T2, T3, T4, . . . . . . . Tn
di mana T ialah sebutan siri tersebut
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NISBAH SEPUNYA
r = T
Tn
n 1
---------- (4)
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CONTOHTentukan nisbah sepunya bagi janjang tersebuta) 1, 3, 9, 27, …
r = r = r= = 3 = 3 =3nisbah sepunya = 3
b) 3p, 6, 12, 24, … r = r = r= = 2p = 2p =2pnisbah sepunya = 2p
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SEBUTAN KE-n
Tn = arn 1
• di mana, Tn = sebutan ke-n
a = sebutan pertama r = nisbah sepunya n = nombor turutan
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1) Cari sebutan ke-7 dan ke-15 untuk siri 4, 12, 36, . . . . . .. Penyelesaian :• diberi sebutan pertama: T1 = a = 4
• sebutan ke-2 : T2 = 12
dari hukum r = T
Tn
n 1
r = T
T2
1
= 12
4
= 3
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dari hukum Tn = arn 1
untuk n = 7, T7 = ar6 = 4 36 = 2916
untuk n = 15, T15 = ar14 = 4 314 = 19131876
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2) Jika laju maksima suatu mesin ialah 600 p.s.m dan laju minima ialah 20 p.s.m. Mesin tersebut perlu dilengkapkan dengan 5 kelajuan. Cari 3 lagi kelajuan yang diperlukan dengan menggunakan kaedah Janjang Geometri.
• Penyelesaian sebutan pertama, T1 = a = 20
sebutan kelima, T5 = 600
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Tn = arn1
T5 = ar4
r4 = T
a5
= 600
20
= 30
r = 301
4
r = 2.34
Sebutan ke- 2, T2 = ar
= 20 2.34
= 46.81 psm
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Sebutan ke- 3, T3 = ar2
= 20 2.342 = 109.54 psm
Sebutan ke- 4, T4 = ar3
= 20 2.343 = 256.37 psm
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HASIL TAMBAH SEBUTAN KE-n
Sn = a r
r
n1
1
di mana Sn = Jumlah sebutan hingga ke-n a = sebutan pertama r = nisbah sepunya n = nombor turutan
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1) Dapatkan jumlah untuk siri 4, 10, 25, . . . . . . hingga sebutan ke-7.
Penyelesaian
r = T
T2
1
= 10
4
= 2.5
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Sn = a r
r
n1
1
S7 = a r
r
1
1
7
= 4 1 2 5
1 2 5
7
.
.
= 1624.94
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1. Find the sum of a) first 6 terms of GP 4, 12, 36,….b) first 7 terms of GP 16, 4, 1,….
2. Given a GP 27,9,3,…. Find the sum from the 4th term to the 9th term of GP.
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Jawapan : 1. a) S6= 1456
b) S7 = 21.33
2. 1.498
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2) Pengeluaran barangan sebuah kilang bertambah sebanyak 8% setiap bulan. Jika pengeluaran pada bulan pertama ialah sebanyak 1500 barangan.
i. Berapa bulankah yang diperlukan untuk menghasilkan sebanyak 16000 barangan. ii. Apakah kadar pengeluaran pada bulan terakhir.
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Penyelesaian : Bahagian (i) sebutan pertama T1 = a = 1500
sebutan ke-2 T2 = 1500 + 1500 8
100 = 1620
sebutan ke-3 T3 = 1620 + 1620 8
100 = 1749.6
siri untuk kadar pengeluaran kilang ini ialah 1500, 1620, 1749.6, . . . . . . . . .
r = T
T2
r = 1620
1500
r = 1.08
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Katakan pada kali ke-n jumlah pengeluaran ialah 16000 dari hukum Sn =
a r
r
n1
1
( 1 r ) Sn = a ( 1 rn ) ( r 1) Sn = a ( rn 1 )
( rn 1) = r S
an 1
rn = r S
an 1
+ 1
1.08n = ( . )108 1 16000
1500
+ 1
1.08n = ( . )108 1 16000
1500
+ 1
1.08n = 1.853
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1.08n = 1.853 ‘log’kan pada kedua-dua belah log (1.08n) = log ( 1.853 )
n = log .
log .
1853
108
n = 8 Untuk menghasilkan barangan sebanyak 16000 memerlukan pengeluaran selama 8 bulan.
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Bahagian (ii)
dari hukum Tn = arn 1 untuk n = 8 T8 = ar7 = 1500 1.087 = 2570.74 2570 Kadar pengeluaran pada bulan terakhir ini ialah 2570 barangan.
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HASIL TAMBAH KETAKTERHINGGAAN
= , |r|< 1
CONTOH :a) 12, 4, = = 18
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b) For a geometric progression, the first term is 18 and the 4th term is . Find i) the common ratio ii) the sum to infinityi) = ii) = = = 54 18= = r =
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c) Express each of the following recurring decimals as a fraction in its simplest form. i) 0.6666… ii) 0.
i) 0.6666… = 0.6 + 0.06 + 0.006 + 0.0006 + … a = 0.6 r = = 0.1 = = = Therfore, 0.6666… =
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ii) 0. = 0.121212… = 0.12 + 0.0012 + 0.000012 + … a = 0.12 r = = 0.01
= = = = = = therefore, 0. =