2.SUKU KETIGA DERET GEOMETRI Vn =48 TENTUKAN JUMBLAH 6 SUKU PERTAMA
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NO NGASAL
Nomor 1
2, 4, 8,...
a = 2
r = 4 ÷ 2 = 2
- Jumlah 5 suku pertama
[tex]S_5 = \frac{ {a(r}^{n} - 1 )}{r - 1} \\ \\ = \frac{2 \times ( {2}^{5} - 1) }{2 - 1} \\ \\ = \frac{2 \times (32 - 1)}{1} \\ \\ = 2 \times 31 \\ \\ = \bold{ 62}[/tex]
[tex] \\ [/tex]
Nomor 2
- Menentukan rasio (r)
[tex]\frac{U_6}{U_3} = \frac{384}{48} \\ \\ \frac{ { \bcancel{a}r}^{5} }{ { \bcancel{a}r}^{2} } = \frac{384}{48} \\ \\ {r}^{5 - 2} = 8 \\ \\ {r}^{3} = 8 \\ \\ r = \sqrt[3]{8} \\ \\ r = 2[/tex]
- Menentukan suku pertama (a)
[tex]U_3 = {ar}^{2} \\48 = a \times {2}^{2} \\ 48 = a \times 4 \\ a = 48 \div 4 \\ a = 12[/tex]
- Menentukan jumlah 6 suku pertama (S6)
[tex]S_n = \frac{ {a(r}^{n} - 1) }{r - 1} \\ \\ S_6 = \frac{12 \times ( {2}^{6} - 1) }{2 - 1} \\ \\ S_6 = \frac{12 \times (64 - 1)}{1} \\ \\ S_6 = 12 \times 63 \\ \\ S_6 =\bold{ 756}[/tex]
Jawaban:
mencari nilai rasio
u6 = a.r^5 = 384
u3 = a. r^2 = 48
-------------------------- :
r^3 = 8
r = akar pangkat tiga dari 8
r = 2
mencari nilai a
u3 = .
a. r^2 = 48
a . (2)^2 = 48
a. 4 = 48
a = 12
jumlah 5 suku pertama
S5 = (a.( r^n - 1)) : (r - 1)
s5 = (12. (2^5 - 1) : (5 - 1)
s5 = (12 . (32 -1) : 4
s5 = (12 . 31) : 4
s5 = 372 : 4
s5 = 93
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