the sum of the first and ninth term of an arithmetic progression is 24.find the sum of the first nine terms of this progression
1. the sum of the first and ninth term of an arithmetic progression is 24.find the sum of the first nine terms of this progression
semoga membantu. maaf, jika salah
2. The fourth term of an AP is 1 and the sum of the first 8 terms is 24. Find the sum of the first 3 terms of the progression. Please help me. Thx.
Let a be the first term and b be the common difference, then
[tex]\begin{array}{rcl}u_4&=&a+3b\\a+3b&+&1\\a&=&1-3b\\\\S_8&=&\frac{8}{2}(2a+7b)\\24&=&4(2(1-3b)+7b)\\6&=&2-6b+7b\\b&=&4\\\\S_3&=&\frac{3}{2}(2a+2b)\\&=&\frac{3}{2}(2(1-3b)+2b)\\&=&\frac{3}{2}(2-6b+2b)\\&=&\frac{3}{2}(2-4b)\\&=&\frac{3}{2}(2-4\times4)\\&=&-21\end{array}[/tex]
3. in arithmetic sequence, the sum of the first ten terms is 125 and the third term is 5. Find the first term, the common difference and the sum of the first 15 terms
Arithmetic Sequence
• The nth term (Un)
Un = a + (n - 1) b
• The sum of the first n terms (Sn)
Sn = n/2 × (a + Un)
Sn = n/2 × (2a + (n - 1) b)
a = the first term
b = the common difference
==================================
S₁₀ = 125
U₃ = 5
a = ?
b = ?
S₁₅ = ?
S₁₀ = 125
10/2 × (2a + (10 - 1) b) = 125
5 × (2a + 9b) = 125
2a + 9b = 25 ... eq (1)
U₃ = 5
a + (3 - 1) b = 5
a + 2b = 5 ... eq (2)
To find a and b, use the elimination/substitution
▪︎Find the first term
Eliminate variable b to find a
2a + 9b = 25 (×2)
a + 2b = 5 (×9)
4a + 18b = 50
9a + 18b = 45
___________ -
-5a = 5
a = -1
The first term is -1
▪︎Find the common difference
To find the common difference, substitute a = -1 to eq (2)
a + 2b = 5
-1 + 2b = 5
2b = 5 + 1
2b = 6
b = 3
The common difference is 3
▪︎Find the sum of the first 15 terms
S₁₅ = 15/2 × [2(-1) + (15 - 1)(3)]
S₁₅ = 15/2 × [-2 + (14)(3)]
S₁₅ = 15/2 × (-2 + 42)
S₁₅ = 15/2 × 40
S₁₅ = 15 × 20
S₁₅ = 300
The sum of the first 15 terms is 300.
Hope it helps.
4. the first term of a progression is 16 and the second term is 24. find the sum of the first eight terms, given that the progression is arithmetic
Sn = n/2 (2a+(n-1)b)
S8 = 8/2 (2 x 16 + (8-1) 8)
S8 = 4 (32 + 56)
S8 = 4 (88)
S8 = 352
Jumlah 8 suku pertama adalah 352
Penjelasan dengan langkah-langkahSoal tersebut membahas tentang deret aritmatika. Deret aritmatika atau dikenal sebagai barisan dan deret hitung adalah barisan yang mempunyai pola tertentu, yakni selisih dua suku berturutan sama dan tetap. Barisan merupakan kelompok angka atau bilangan yang berurutan, sedangkan deret merupakan jumlah dari suku-suku pada barisan.
Diketahui:
a = suku pertama = 16b = beda = 8n = banyak suku = 8Ditanya:
S8 = Jumlah 8 suku pertama = ?Jawab:
Sn = n/2 (2a+(n-1)b)
S8 = 8/2 (2 x 16 + (8-1) 8)
S8 = 4 (32 + 56)
S8 = 4 (88)
S8 = 352
Sehingga jumlah 8 suku pertama adalah 352
Pelajari lebih lanjutyuk belajar tentang barisan dan deret di sini https://brainly.co.id/tugas/1381755
#BelajarBersamaBrainly
5. The sum of the first 25 terms of an arithmetic sequence with 50 terms is 400, meanwhile the sum of the next 25 terms is 2275. Then the first term of the sequence is... . (A) —50 (D) 10 (B) —20 (E) 20 (C) 5
Jawaban:
A) —50
Penjelasan:
karena The sum of the first 25 terms of an arithmetic sequence with 50 terms is 400, meanwhile the sum of the next 25 terms is 2275. Then the first term
6. find the sum of the first 3 multiple of 7
Jawaban:
temukan jumlah dari 3 kelipatan 7 pertama
Jawaban:
temukan jumlah dari 3 kelipatan 7 pertama
7. let Sn, be the sum of the first n terms of the airthmetic series 2+4+6... (a) find S4
2 + 4 + 6 + ... + Un
⇒ a = 2, b = 4 - 2
maka
Sn = (n/2)(2a + (n - 1)b)
⇒ Sn = (n/2)(2(2) + (n - 1)2)
⇒ Sn = (n/2)(4 + 2n - 2)
⇒ Sn = (n/2)(2n + 2)
⇒ Sn = (n/2) · 2(n + 1)
⇒ Sn = (n)(n + 1)
Jadi,
S4 = 4(4 + 1)
⇒ S4 = 4 · 5
⇒ S4 = 20
Terimakasih semoga membantu2 + 4 + 6 + ...
a = 2 (first term)
d = 2 (difference of the term)
S4 = (n/2)(2a + (n - 1)d)
= (4/2)(2.2 + (4 - 1)2)
= 20
8. the first three terms in geometric progression are (7x-7),(2x + 1),and (x-3) where x is a positive integer.find the value of x and find the sum of the first 5 terms of the progression
Kelas 8 Matematika
Bab Barisan dan Deret Bilangan
U1 = 7x - 7
U2 = 2x + 1
U3 = x - 3
(U2)² = U1 . U3
(2x + 1)² = (7x - 7) (x - 3)
4x² + 2 . 2x + 1 + 1² = 7x² - 21x - 7x + 21
4x² + 4x + 1 = 7x² - 28x + 21
7x² - 4x² - 28x - 4x + 21 - 1 = 0
3x² - 32x + 20 = 0
(3x - 2) (x - 10) = 0
3x - 2 = 0
3x = 2
x = 2/3
x - 10 = 0
x = 10
x = 10
U1 = 7x - 7
U1 = 7 . 10 - 7
U1 = 63 → a
U2 = 2x + 1
U2 = 2 . 10 + 1
U2 = 21
U3 = x - 3
U3 = 10 - 3
U3 = 7
r = U2 / U1
r = 21 / 63
r = 1/3
Sn = a . (1 - r^n)/(1 - r)
S5 = 63 . (1 - (1/3)^5)/(3/3 - 1/3)
S5 = 63 . (243/243 - 1/243) / (2/3)
S5 = 63 . 242/243 . 3/2
S5 = 7 . 121 /9
S5 = 847/9
S5 = 94 1/9
9. 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.
10. If (m+ 1), (2m - 7), and (m + 7) are the 1st,2nd, and 3nd terms of an Arithmetic Sequeace, respectively A. Find m B. Find the formula for finding the nth term. C. Find the 7th term D. Find the sum of the first 10 terms
Jawaban:
C
Penjelasan dengan langkah-langkah:
ngak tau (づ｡◕‿‿◕｡)づ(o´･_･)っ
Jawaban:
C. Find the 7th term
Penjelasan dengan langkah-langkah:
Maaf Kalau salah...
jadikan jawaban terbaik...
jangan lupa follow yaa..
11. If the sum of the first 10 terms and the sum of the first 100 terms of a given arithmetic progression are 100 and 10, respectively, then the sum of the first 110 terms is... . (A) —140 (B) —130 (C) —120 (D) —110 (E) —100
Jawaban:
(C) —120
Penjelasan:
because 10+100+100+10=-120
12. Find the sum of this geometric series 3 – 6 + 12 - ... to 7th terms
3 - 6 + 12 - 2 = 7
So the answer is 2
13. Find the first two terms of an arithmetic sequence if the sixth term is 21 and the sum of the first seventeen terms is 0
Jawaban:
56 and 49
Penjelasan dengan langkah-langkah:
u6 = a + 5b = 21
s17 = 0
u9 = a + 8b = 0
-3b = 21
b = -7
a = 56
so the first two terms are 56 and 49
14. The first two terms of an arithmetic progression are 15 and 19 respectively. The firsttwo terms of a second arithmetic progression are 420 and 415 respectively. The twoprogressions have the same sum of the first n terms. Find the value of n.
Jawaban:
n = 91
Penjelasan dengan langkah-langkah:
Jumlah n suku sama sehingga
Sn = ½ n (2.15 + 4(n-1)) = ½ n (2.420 + -5(n-1))
30 + 4n - 4 = 840 -5n +5
4n + 5n = 845-26
9n = 819
n = 91
15. Find the sum of the positive terms of the arithmetic sequence 85, 78, 71,
Jawaban:
78,85,71
Penjelasan dengan langkah-langkah:
maaf kalo salah
16. The sum of the first 20 terms 2+8+14+20+..., is ...
Aritmatika
U1 = a = 2
b = 8 - 2 = 6
Sn = n/2 x (2a + (n - 1) b)
S20 = 20/2 x (2 x 2 + 19 x 6)
S20 = 10 x 118
S20 = 1.180
17. The sum of the first two terms 2 + 8 + 14 + 20 + ..., is ....
the sum of the first two terms = U1 + U2 = 2 + 8 = 10
18. The sum of the first four terms in a geometricsequence is 30 and the sum to infinity is 32. The firstthree terms of the sequence are ....
Jawaban:
Jumlah dari empat suku pertama dalam geometri
urutannya adalah 30 dan jumlah hingga tak terhingga adalah 32. Yang pertama
tiga suku urutan tersebut adalah ....
19. 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
20. The sum of two numbers is 20 and the sum of their squares is 272. Find the firstnumber
Jawab:
4 or 16
Penjelasan dengan langkah-langkah:
Let the numbers be x and y
x + y = 20
sum of their squares is 272 so x^2 + y^2 = 272
Find the first number,
We know that x^2 + y^2 = (x + y)^2 - 2xy
so 272 = 20^2 - 2xy
272 = 400 - 2xy
2xy = 128
xy = 64
We know that x + y = 20 and xy = 64, so we can make quadratic equation,
let a1 = x and a2 = y
a^2 - 20a + 64 = 0
(a - 16)(a - 4) = 0
a = 16 or a = 4
So the first number can be 4 or 16
If the first number less than the second, the answer is 4
Semoga terbantu :)
IG: @djie.jemmy
Jawab:
4 or 16Penjelasan dengan langkah-langkah:
If the first number is [tex]x[/tex] and the second one is [tex]y[/tex],
we can write:
[tex]x + y = 20[/tex]
[tex]x^{2} + y^{2} = 272[/tex]
Then
[tex]x = 20 - y[/tex]
Substitute [tex]x = 20 - y[/tex] to [tex]x^{2} + y^{2} = 272[/tex]
[tex](20 - y)^{2} + y^{2} = 272[/tex]
[tex]400 - 40y + y^{2} + y^{2} = 272[/tex]
[tex]2y^{2} - 20y + 128 = 0[/tex]
Factorize the equation and we get:
[tex](y-16)(y-4) = 0[/tex]
[tex]y = 16[/tex] ∨[tex]y = 4[/tex]
If we substitute [tex]y = 16[/tex] to [tex]x = 20 - y[/tex], we get [tex]x = 4[/tex]
So, we can conclude that both numbers are 4 and 16.
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