I have a geometric series problem . please help me to solve it Problem: The third term of a geometric sequence is 3 and the sixth term is 1/9. Find the first term.
1. I have a geometric series problem . please help me to solve it Problem: The third term of a geometric sequence is 3 and the sixth term is 1/9. Find the first term.
The 1st term is 27.
@wello1
☺
2. The second term of a geometric series with a positive ratio is 10 and the 6th term is 160 the sum of the first 10 terms of the series is?
[tex] {r}^{4} = \frac{u6}{u2} = \frac{160}{10} = 16 \\ r = \sqrt[4]{16} = 2 \\a = \frac{u2}{r} = \frac{10}{2} = 5 \\ sn = \frac{a( {r}^{n} - 1)}{r - 1} \\ s10 = \frac{5( {2}^{10} - 1)}{2 - 1} \\ s10 = \frac{5(1024 - 1)}{1} \\ s10 = 5(1023) \\ s10 = 5115[/tex]
3. The sum of a given infinite geometric series is 200, and the common ratio is 0,15. What is the second term in this series?A. 25,5B. 82,5C. 30D. 170
Jawaban:
The second term in this series is 25,5. (A)
Penjelasan:
Deret Geometri Takhingga
[tex]\boxed{~S\infty~=~\frac{a}{(1-r)} ~}[/tex]
Diketahui:
Deret geometri tak hingga
The sum of a given infinite geometric series is 200.
S∞ = 200.
The common ratio is 0,15.
r = 0,15
Ditanya:
What is the second term in this series?
Suku ke-2 dari barisan?
Jawab:
Langkah pertama kita cari suku pertama terlebih dahulu.
S∞ = [tex]\frac{a}{1-r}[/tex]
200 = [tex]\frac{a}{1-0,15}[/tex]
200 = [tex]\frac{a}{0,85}[/tex]
a = 200 x 0,85
a = 170
Langkah selanjutnya kita cari suku kedua dari barisan
U₂ = a.r
= 170 x 0,15
= 25,5
The second term in this series is 25,5. (A)
4. In a series geometry, the sum of the first nine series is 511, if the ratio is 2, determine the 2nd and the 4th term
2nd = 2
4th = 8
Hope you get a satisfactory score
5. The rule to write any term in a sequences is T = 5n - 2 Work out the 30th term
Jawaban:
148
Penjelasan dengan langkah-langkah:
30th term, berarti suku ke-30. artinya, n = 30
T = 5n - 2
= 5(30) - 2
= 150-2
= 148
6. Find the missing term or terms in each arithmetic sequence. a. ..., - 20, ____, ____, ____, 0, ....
Jawaban:
-15, -10, -5
semoga membantu
terima kasih
Jawaban:
-25,-20,-15,-10,-5,0,5
7. In a geometric sequence, the sum of the first three terms is 76/45 and the sum of the next three terms is 608/1215 . Find the common ratio and the first term of the sequence.
Penjelasan dengan langkah-langkah:
[tex]U_1 + U_2 + U_3 = \frac{76}{45}[/tex]
[tex]a+ar+ar^2 = \frac{76}{45}\\[/tex]
[tex]U_4 + U_5 + U_6 = \frac{608}{1215}[/tex]
[tex]ar^3+ar^4+ar^5= \frac{608}{1215}[/tex]
[tex]r^3(a+ar+ar^2)= \frac{608}{1215}[/tex]
[tex]r^3 \cdot \frac{76}{45}= \frac{608}{1215}[/tex]
[tex]r^3 = \frac{608}{1215} \cdot \frac{45}{76}[/tex]
[tex]r^3 = \frac{4\cdot 152}{5\cdot 243} \cdot \frac{5\cdot 9}{4\cdot 19}[/tex]
[tex]r^3 = \frac{152}{243} \cdot \frac{9}{19}[/tex]
[tex]r^3 = \frac{8\cdot 19}{9\cdot 27} \cdot \frac{9}{19}[/tex]
[tex]r^3 = \frac{8}{27} [/tex]
[tex]r = \frac{2}{3}\\\\[/tex]
[tex]a+ar+ar^2 = \frac{76}{45}[/tex]
[tex]a+\frac{2}{3}a+\frac{4}{9}a = \frac{76}{45}[/tex]
[tex]\frac{9}{9}a+\frac{6}{9}a+\frac{4}{9}a = \frac{76}{45}[/tex]
[tex]\frac{19}{9}a = \frac{76}{45}[/tex]
[tex]a= \frac{76}{45}\cdot \frac{9}{19} [/tex]
[tex]a= \frac{4\cdot 19}{9 \cdot 5}\cdot \frac{9}{19} [/tex]
[tex]a= \frac{4}{5}\\[/tex]
so, the ratio is 2/3 and the first term is 4/5
8. What is the value of first term and common ratio of the geometric sequence if the third term is 96 and the fifth term is 1,536?
Jawaban:
cara dan jawabannya seperti di foto ya.
semangat belajar
semoga membantu
#terbaik5
9. Geometric Series&Sequences 1. Find U4,S4 & S∞ for the series below. 144 + 48 + 16 + ... 2. For a geometric sequence with u3 = 24 and u6 = 3, find S∞. 3. A geometric series has a common ratio of 0.4 and a sum to infinity of 250. Find the first term.
Jawaban:
mohon maaf saya tidak bisa menjawab soal kamu mohon maaf???
10. in the sum of the first term n of arithmetic series in defined by s6= 2n^+3n then the difference of the series is...a. 2b. 3c. 4d. 5e. 6tolong ya
Jawab:
Penjelasan dengan langkah-langkah:
11. in a geometric sequence,T1= -3, the ratio is 4 and the last term is -3072. how many terms are there is this sequence?
Jawaban:
6
Penjelasan dengan langkah-langkah:
first term = -3
ratio (r) = 4
n term = -3072.
number of terms (n) = ....
Un = a .r^n-1
-3072 = -3 .(4)^n-1
1024 = 4^n-1
4⁵ = 4 ^n-1
n-1 = 5
n = 6.
the numbers of terms = 6
12. The third term of a geometric progression is -108 and the sixth term is 32. Find (a) the common ratio and first term. [6 marks] (b) [2 marks] the sum of the first 20th term.
Jawab:
(a) Common ratio, r = $\frac{32}{-108} = -\frac{1}{3}$
First term, a = -108
(b) Sum of the first 20 terms, S$_{20}$ = $\frac{a\left(1-r^{20}\right)}{1-r}$
= $\frac{-108\left(1-(-\frac{1}{3})^{20}\right)}{1-(-\frac{1}{3})}$
= $\frac{-108\left(1-\frac{1}{3^{20}}\right)}{\frac{4}{3}}$
= $\frac{-432\left(1-\frac{1}{3^{20}}\right)}{4}$
= $-108\left(3^{19}-1\right)$
= $-108\left(3^{19}\right) + 108$
= $-3245056 + 108$
= -3244948
13. Find the sum of arithmetic series where the last term is 41, the girst term is 3 and the differences is 2
Un = 41
a = 3
b = 2
Un = 3 + (n - 1) 2 = 41
3 + 2n - 2 = 41
2n + 1 = 41
2n = 40
n = 20S20 = 20/2 (6 + 19(2)
S20 = 10 ( 6 + 38) = 44014. find the sum of the terms of an infinite geometric sequence whose first term is 4 and common ratio ⅕
" Barisan Geometri "
__________
>>>Diketahui:
a = 4
r = ⅕
________
>>> S∞ = ....
___________
[tex] \sf S_{ \infty } = \frac{a}{1 - r} [/tex]
[tex] \sf S_{ \infty } = \frac{4}{1 - \frac{1}{5} } [/tex]
[tex] \sf S_{∞} = \frac{4}{ \frac{4}{5} } [/tex]
[tex] \sf\to 4 \div \frac{ 4}{5} \\ \sf \to \cancel{4} \times \frac{5}{ \cancel{4}} [/tex]
[tex] \boxed{ \sf S_{∞} = 5}[/tex]
____________
CMIIW
Ciyo.
15. The formula of n-term from a number sequence is Un = =n(n − 3). The 8th term is ....2.En in
Penjelasan dengan langkah-langkah:
hshdvdhshdhdhdhdudude77e
U8 = 40
Penjelasan dengan langkah-langkah:
Un = n(n − 3)
U8 = 8 (8-3)
= 8 x 5
= 40
16. each ___ is stored in a column
eachother is store in a column
17. In a geometric series, e^(m+1),(m + 1)e^(m+1) and (4m + 1)e^(m+1) are consecutive terms. a) Calculate the value of m and the common ratio.b) Hence, if the third term of the series is e^(m+1), find the sum of the first 6 terms of the series.
a. Nilai m adalah 0 atau 2 dan r adalah 1 atau 3.
b. Jumlah 6 suku pertama adalah 6e atau [tex]\boldsymbol{\frac{364}{9}e^3 }[/tex].
PEMBAHASANBarisan geometri adalah suatu barisan bilangan dimana bilangan yang berurutan memiliki rasio atau perbandingan yang tetap. Rumus rumus yang terdapat pada deret geometri adalah sebagai berikut.
[tex]u_n=ar^{n-1}[/tex]
[tex]\displaystyle{r=\frac{u_n}{u_{n-1}} }[/tex]
[tex]S_n=\frac{a(r^n-1)}{r-1},~untuk~r> 1[/tex]
[tex]S_n=\frac{a(1-r^n)}{1-r},~untuk~r< 1[/tex]
dengan :
a = suku pertama
r = rasio
[tex]u_n=[/tex] suku ke-n
[tex]S_n=[/tex] jumlah suku ke-n
.
DIKETAHUI[tex]e^{m+1},~(m+1)e^{m+1},~dan~(4m+1)e^{m+1}[/tex] merupakan suku suku berurutan pada barisan geometri.
.
DITANYAa. Tentukan nilai m dan rasionya.
b. Tentukan jumlah 6 suku pertamanya jika suku ke-3 = [tex]e^{m+1}[/tex].
.
PENYELESAIANSoal a.
Pada suku yang berurutan berlaku :
[tex]r=r[/tex]
[tex]\frac{(m+1)e^{m+1}}{e^{m+1}}=\frac{(4m+1)e^{m+1}}{(m+1)e^{m+1}}[/tex]
[tex](m+1)=\frac{(4m+1)}{(m+1)}~~~...kali~silang[/tex]
[tex](m+1)^2=4m+1[/tex]
[tex]m^2+2m+1=4m+1[/tex]
[tex]m^2-2m=0[/tex]
[tex]m(m-2)=0[/tex]
[tex]m=0~atau~m=2[/tex]
.
Mencari rasio barisan.
[tex]r=\frac{u_n}{u_{n-1}}[/tex]
[tex]r=\frac{(m+1)e^{m+1}}{e^{m+1}}[/tex]
[tex]r=m+1[/tex]
.
Untuk m = 0 :
[tex]r=0+1[/tex]
[tex]r=1[/tex]
.
Untuk m = 2 :
[tex]r=2+1[/tex]
[tex]r=3[/tex]
.
Soal b.
Untuk m = 0 dan r = 1 :
[tex]u_3=e^{m+1}[/tex]
[tex]ar^{3-1}=e^{0+1}[/tex]
[tex]ar^2=e[/tex]
[tex]a(1)^2=e[/tex]
[tex]a=e[/tex]
.
Maka :
[tex]S_6=u_1+u_2+u_3+u_4+u_5+u_6[/tex]
[tex]S_6=e+e+e+e+e+e[/tex]
[tex]S_6=6e[/tex]
.
.
Untuk m = 2 dan r =3 :
[tex]u_3=e^{m+1}[/tex]
[tex]ar^{3-1}=e^{2+1}[/tex]
[tex]ar^2=e^3[/tex]
[tex]a(3^2)=e^{3}[/tex]
[tex]a=\frac{e^3}{9}[/tex]
.
Maka :
[tex]S_6=\frac{a(r^n-1)}{r-1}[/tex]
[tex]S_6=\frac{\frac{e^3}{9}((3)^6-1)}{3-1}[/tex]
[tex]S_6=\frac{\frac{e^3}{9}(728)}{2}[/tex]
[tex]S_6=\frac{364}{9}e^3[/tex]
.
KESIMPULANa. Nilai m adalah 0 atau 2 dan r adalah 1 atau 3.
b. Jumlah 6 suku pertama adalah 6e atau [tex]\boldsymbol{\frac{364}{9}e^3 }[/tex].
.
PELAJARI LEBIH LANJUTMencari jumlah barisan geometri : https://brainly.co.id/tugas/29609114Deret geometri tak hingga : https://brainly.co.id/tugas/29553829Deret geometri : https://brainly.co.id/tugas/24888137.
DETAIL JAWABANKelas : 9
Mapel: Matematika
Bab : Barisan dan Deret Bilangan
Kode Kategorisasi: 9.2.2
Kata Kunci : barisan,geometri, rasio, suku.
18. A geometric sequence has 9 terms. If the third term is 80 and the last term is 327680, find the common ratio of the geometric sequence.
[tex]u9 = u3 \times {r}^{6} \\ 327680 = 80 \times {r}^{6} \\ \frac{327680}{80} = {r}^{6} \\ {r}^{6} = 4096 \\ r = \sqrt[6]{4096} \\ r = 4[/tex]
19. the sum of the first n term of a series is given by sn = 2n(n 3) show that the term of the series form an aritmathic progression
Jawaban:
Note:
It's not clear whether it's [tex]S_n=2n(n-3)[/tex] or [tex]S_n=2n(n+3)[/tex].
So, there will be 2 cases.
Barisan Aritmetika (Arithmetic Progression)Case 1: [tex]S_n=2n(n-3)[/tex]
Finding the [tex]a_n[/tex] formula
[tex]\large\text{$\begin{aligned}S_n&=2n(n-3)\\a_n&=S_n-S_{n-1}\\&=2n(n-3)-2(n-1)(n-1-3)\\&=2n(n-3)-2(n-1)(n-4)\\&=2n(n-3)-2n(n-4)+2n-8\\&=2n(n-3-(n-4))+2n-8\\&=2n(n-3-n+4)+2n-8\\&=2n(-3+4)+2n-8\\&=2n(1)+2n-8\\&=2n+2n-8\\&=4n-8\\\therefore\ a_n&=\bf4n-8\end{aligned}$}[/tex]
Thus, the term of the [tex]S_n=2n(n-3)[/tex] form an arithmetic progression of –4, 0, 4, 8, 12, ... .
The difference between each term equals to 4.
Case 2: [tex]S_n=2n(n+3)[/tex]
Finding the [tex]a_n[/tex] formula
[tex]\large\text{$\begin{aligned}S_n&=2n(n+3)\\a_n&=S_n-S_{n-1}\\&=2n(n+3)-2(n-1)(n-1+3)\\&=2n(n+3)-2(n-1)(n+2)\\&=2n(n+3)-2n(n+2)+2n+4\\&=2n(n+3-(n+2))+2n+4\\&=2n(n+3-n-2)+2n+4\\&=2n(3-2)+2n+4\\&=2n(1)+2n+4\\&=2n+2n+4\\&=4n+4\\\therefore\ a_n&=\bf4n+4\end{aligned}$}[/tex]
Thus, the term of the [tex]S_n=2n(n+3)[/tex] form an arithmetic progression of 8, 12, 16, 20, 24, ... .
The difference between each term equals to 4, which is the same as case 1.
20. The first term of a geometric progression is 75 and the third term is 27. Find the possible values for the fourth term
Jawab:
terlampir
Penjelasan dengan langkah-langkah:
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