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.

1. 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]

2. The first term of an arithmetic sequence is 14. The fourth term is 32. Find the common difference.

Answer:

The n-th term of an arithmetic sequence is given by:

Un = a + (n - 1)b

Where a the first term, b the common difference. If U4 = 32 and a = 14 then

32 = 14 + (4 - 1)b

18 = 3b

b = 6

The common difference is 6

3. the first term of an arithmetic progression is 3, the fourth term is 15 and the 16th term is 63, find the common difference of this progression.​

Jawab:

b = difference = 4

Penjelasan dengan langkah-langkah:

U1 = 3, U4 = 15, U16 = 63

U1 = a = 3

U4 = a + 3b

15 = 3 + 3b

3b = 15 - 3 = 12

b = 12/3 = 4

4. 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

5. 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-langkah

Soal 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 = 8

Ditanya:

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 lanjut

yuk belajar tentang barisan dan deret di sini https://brainly.co.id/tugas/1381755

#BelajarBersamaBrainly

6. given the 3rd and 5th term of an arithmetric progression is q + 2p and q+4p respectively find the first term and common difference​

Penjelasan dengan langkah-langkah:

a+ 2b = q+2p

a+ 4b = q+ 4p

--------------------- -

-2b = -2p

b = p → a +2(p) = q+2p

a = q

the first term = q

common difference = p

7. 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

8. 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

9. 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.

10. Find the first three terms of the sequences with the given nth term n(2n + 1) boleh minta penjelasannya =)

Jawab:

Penjelasan dengan langkah-langkah:

Jadi itu disuruh bikin 3 bilangan pertama dari pola bilangan dengan rumus itu.

n=1,n=2,n=3

U1=n(2n+1)

U1=1(2+1)

U1=3

U2=2(2x2+1)

U2=2(4+1)

U2=10

U3=3(2x3+1)

U3=3(6+1)

U3=21

=3,10,21

Moga membantu ^^

11. 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.

12. 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.

13. the first term is 3 and the fourth is 9. find 11th term.

Jawaban:

suku pertama adalah 3 dan suku keempat 9. temukan suku kesebelas.

Un = a + (n-1)b

ket :

Un = suku ke-n

a = suku pertama

b = beda

     b = U_{n} - U_{n-1}b=Un−Un−1

n = banyaknya suku

Unnya 8

14. 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:

15. 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..

16. Tolong Bantu jwb : 1.One term of an arithmetic sequence is T13 = 30. The common diffrence is d = 3/2 a. Find the first Term. b. Write a rule for the nth term. 2. In an arithmetic series, T1 = 5 and T20 = 62. a. Find the common diffrence. 2. Find the value of T15. 3. Find the sum of the first 40 terms. (S40) 3. Consider the arithmetic series 4+7+10+13+16+19+... Find the sum of the first 30 terms. (S30) #Tolong bantu jwb ya gk ada cara juga gpp. Thx Senin di kumpul

3. Arithmetic series
4 + 7 + 10 + 13 + 16 + 19 + ...
a = 4 (first term), d = 3 (difference)
Sn = (n/2)(2a + (n - 1)d)
S10 = (10/2)(2(4) + (10 - 1)3)
= 175

17. consider the sequence 3,6,9,12... A)find the general term of the sequence B)Find the 11th term of the sequence​

Jawaban:

3,6,9,12,...

A) +3 / multiplication 3

B) 3,6,9,12,15,18,21,24,27,30,(33) / 3 × 11 = 33

jawabanny ad 2 ya, yg paling sesuai aj.

semoga membantu:)

18. the difference between the tenth term and the seventh term of an arithmetic sequence is -60.the twelfth term divided by the sixth term is 2.find the first term and the common difference.

U10-U7= -60
a= first term , d= common difference
a+9d - (a+6d) = -60
a+9d -a -6d =-60
3d= -60
d= -20

U12/U6 = 2
U12=2U6
a+11d=2(a+5d)
a+11d=2a+10d
d=a=-20

19. consider this sequence 3,6,12,24,48,find in the term of. ,the formula of the nth term of this sequence​

Jawaban:

the formula was 3.2^n

while n is respective number from 0 to unlimited. but its a round number, not partial one

20. find the nth term of the sequence 3,8,15,24

Polanya itu ditambah terus sama bilangan ganjil
3(+5), 8(+7), 15(+9), 24
Jadi bilangan selanjutnya 24 + 11 = 35