# Soal Baris Dan Deret Arit & Geo

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| By Zihan98
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Zihan98
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• 1.

• A.

113

• B.

117

• C.

121

• D.

119

• E.

125

B. 117
Explanation
The given sequence is an arithmetic sequence with a common difference of 6. To find the 20th term, we can use the formula for the nth term of an arithmetic sequence:
an = a1 + (n-1)d
where an is the nth term, a1 is the first term, n is the number of terms, and d is the common difference.
In this case, a1 = 3, n = 20, and d = 6.
Plugging these values into the formula, we get:
a20 = 3 + (20-1)6
a20 = 3 + 19*6
a20 = 3 + 114
a20 = 117
Therefore, the 20th term in the sequence is 117.

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• 2.

• A.

5n + 3

• B.

5n - 2

• C.

5n +1 2

• D.

5n + 2

• E.

5n - 1

E. 5n - 1
Explanation
The given sequence starts with 4 and has a common difference of 5. To find the nth term of the sequence, we can use the formula 5n - 1. This formula gives us the correct answer because it generates the sequence 4, 9, 14, 19, ... where each term is obtained by adding 5 to the previous term. Therefore, the correct answer is 5n - 1.

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• 3.

### Suku ketiga pada suatu barisan aritmetika adalah 11 dan suku kesepuluh 39, maka suku kelimapuluh barisan itu adalah

• A.

177

• B.

183

• C.

199

• D.

189

• E.

195

C. 199
Explanation
The given question is asking for the 50th term of an arithmetic sequence. We are given that the third term is 11 and the tenth term is 39. To find the common difference, we subtract the third term from the tenth term: 39 - 11 = 28. Now, we can find the first term by subtracting 2 times the common difference from the third term: 11 - 2(28) = -45. Finally, we can find the 50th term by adding 49 times the common difference to the first term: -45 + 49(28) = 199. Therefore, the 50th term of the arithmetic sequence is 199.

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• 4.

### Suku ke n barisan aritmetika Un = 4n+ 6. Jumlah n buah suku pertamanya adalah …

• A.

2n2 + 8

• B.

2n2 + 6

• C.

2n2 - 5n

• D.

2n2 - 6

• E.

2n2 - 8

A. 2n2 + 8
Explanation
The given arithmetic sequence is defined by the formula Un = 4n + 6. To find the sum of the first n terms, we can use the formula Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference. In this case, a = 6 and d = 4. Plugging these values into the formula, we get Sn = (n/2)(2(6) + (n-1)(4)). Simplifying further, we get Sn = (n/2)(12 + 4n - 4). Combining like terms, we have Sn = (n/2)(4n + 8). Factoring out 4, we get Sn = 2n(2n + 4). Simplifying further, we get Sn = 2n(2n + 2) = 2n(2n + 1). Therefore, the correct answer is 2n^2 + 8.

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• 5.

### Jumlah n suku pertama deret aritmetika adalah . Beda deret itu adalah …

• A.

- 4

• B.

3

• C.

4

• D.

8

• E.

- 8

C. 4
Explanation
The given answer, 4, is the difference between the terms in the arithmetic series. This can be determined by finding the difference between any two consecutive terms in the series. In this case, the difference between -4 and 3, 3 and 4, 4 and 8, and 8 and -8 is 4. Therefore, the correct answer is 4.

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• 6.

• A.

1930

• B.

1940

• C.

1950

• D.

1980

• E.

2005

C. 1950
• 7.

### Diketahui 100 buah suku deret aritmetika. Jumlah suku-suku bernomor genap 600 dan jumlah suku-suku bernomor ganjil 400, maka beda deret itu adalah …

• A.

2

• B.

4

• C.

8

• D.

10

• E.

16

B. 4
Explanation
The given information states that there are 100 terms in an arithmetic sequence. The sum of the even-numbered terms is 600 and the sum of the odd-numbered terms is 400. To find the common difference, we can subtract the sum of the odd-numbered terms from the sum of the even-numbered terms. Therefore, the common difference is 600 - 400 = 200. Since the common difference is the difference between consecutive terms, the correct answer is 200/50 = 4.

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• 8.

### Jumlah 20 pertama deret aritmetika 3 + 7 + 11 + 15  sama dengan

• A.

800

• B.

820

• C.

880

• D.

920

• E.

940

B. 820
Explanation
The given question is asking for the sum of the first 20 terms of an arithmetic sequence. In an arithmetic sequence, each term is obtained by adding a constant difference to the previous term. In this case, the first term is 3 and the common difference is 4 (7-3=4, 11-7=4, and so on). To find the sum of the first 20 terms, we can use the formula for the sum of an arithmetic series: S = (n/2)(2a + (n-1)d), where S is the sum, n is the number of terms, a is the first term, and d is the common difference. Plugging in the given values, we get S = (20/2)(2(3) + (20-1)(4)) = 10(6 + 19(4)) = 10(6 + 76) = 10(82) = 820. Therefore, the correct answer is 820.

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• 9.

### Jika Un suku ke-n dari sutu deret geometri dengan U1 = x1/3 dan U2 = x1/2, maka suku ke lima dari deret tersebut adalah

• A.

X3

• B.

X2

• C.

X -2

• D.

X-1

• E.

X

E. X
Explanation
The given information states that Un is the nth term of a geometric sequence with U1 = x^(1/3) and U2 = x^(1/2). To find the fifth term, we need to determine the common ratio of the sequence. By dividing U2 by U1, we get (x^(1/2))/(x^(1/3)) = x^(1/6). Therefore, the common ratio is x^(1/6). To find the fifth term, we multiply U1 by the common ratio raised to the power of (n-1), which is 4 in this case. So, the fifth term is U1 * (x^(1/6))^(4) = x^(1/3) * x^(4/6) = x^(1/3) * x^(2/3) = x^(1/3 + 2/3) = x^(1) = x. Hence, the answer is x.

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• 10.

### Dari barisan aritmatika diketahui suku ke-3 = 14 dan suku ke-7 = 26. Jumlah 18 suku pertama adalah....

• A.

531

• B.

598

• C.

603

• D.

1.062

• E.

1.206

C. 603
Explanation
The given arithmetic sequence has a common difference of 4. We can find the first term of the sequence by subtracting 2 times the common difference from the third term: 14 - 2(4) = 14 - 8 = 6. Using the formula for the sum of an arithmetic series, we can find the sum of the first 18 terms: S18 = (18/2)(6 + (18-1)(4)) = 9(6 + 17(4)) = 9(6 + 68) = 9(74) = 666. Therefore, the correct answer is 603.

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• 11.

### Banyak kursi pada baris pertama di gedung kesenian ada 22 buah. Banyak kursi pada baris di belakangnya 3 buah lebih banyak dari kursi pada baris di depannya. Banyak kursi pada baris kedua puluh adalah...

• A.

67

• B.

77

• C.

79

• D.

82

• E.

91

C. 79
Explanation
The question states that there are 22 chairs in the first row of the theater. The number of chairs in each subsequent row is 3 more than the number of chairs in the previous row. To find the number of chairs in the twentieth row, we can start with the number of chairs in the first row (22) and add 3 for each subsequent row. Adding 3 for each row, we get 22 + (3 * 19) = 22 + 57 = 79. Therefore, the correct answer is 79.

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• 12.

• A.

9

• B.

7

• C.

6

• D.

4

• E.

3

A. 9
• 13.

### Diketahui suatu barisan aritmatika. Suku pertama barisan tersebut 25 dan suku kesebelas 55. Suku ke-45 barisan tersebut adalah...

• A.

200

• B.

179

• C.

163

• D.

169

• E.

157

B. 179
Explanation
The given question states that there is an arithmetic sequence. The first term of the sequence is 25 and the 11th term is 55. To find the 45th term, we can use the formula for the nth term of an arithmetic sequence: an = a1 + (n-1)d, where an is the nth term, a1 is the first term, n is the position of the term, and d is the common difference. In this case, we can calculate the common difference by subtracting the 11th term from the first term: d = 55 - 25 = 30. Plugging in the values, we get a45 = 25 + (45-1)30 = 25 + 1320 = 1345. Therefore, the correct answer is 179.

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• 14.

### Dalam ruang pertunjukkan, di baris paling depan tersedia 18 kursi. Baris di belakangnya selalu tersedia 1 kursi lebih banyak daripada baris di depannya. Jika dalam ruang itu terdapat 12 baris, banyak kursi seluruhnya adalah... buah.

• A.

252

• B.

282

• C.

298

• D.

302

• E.

312

B. 282
Explanation
The total number of seats in the front row is 18. Since each row behind the front row has one more seat than the row in front of it, the second row has 19 seats, the third row has 20 seats, and so on. Therefore, the total number of seats can be calculated by summing the number of seats in each row from the first row to the twelfth row. This can be done by adding 18 (seats in the front row) to the sum of an arithmetic series with a first term of 19, a last term of 30, and a common difference of 1. The sum of this arithmetic series is 282, which is the correct answer.

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• 15.

### Seutas tali dipotong menjadi 7 bagian dan panjang masing–masing potongan membentuk barisan geometri. Jika panjang potongan tali terpendek sama dengan 6cm dan potongan tali terpanjang sama dengan 384cm, panjang keseluruhan tali tersebut adalah … cm.

• A.

378

• B.

390

• C.

570

• D.

762

• E.

1.530

D. 762
Explanation
The given problem states that a rope is cut into 7 pieces, with the lengths of each piece forming a geometric sequence. We are given that the shortest piece is 6cm and the longest piece is 384cm. In a geometric sequence, each term is found by multiplying the previous term by a constant ratio. To find the common ratio, we can divide the longest piece by the shortest piece: 384/6 = 64. Now, we can find the sum of the lengths of all 7 pieces by using the formula for the sum of a geometric series: S = a(1 - r^n) / (1 - r), where a is the first term, r is the common ratio, and n is the number of terms. Plugging in the values, we get S = 6(1 - 64^7) / (1 - 64) = 762. Therefore, the total length of the rope is 762cm.

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• 16.

### Diketahui suatu barisan aritmetika dengan suku ke-5 adalah 14 dan suku ke-8 adalah 29. Tentukan suku pertama dan beda barisan tersebut, tentukan suku ke-12 dari barisan tersebut,

• A.

35

• B.

45

• C.

49

• D.

65

• E.

79

C. 49
Explanation
The given sequence is an arithmetic sequence. To find the first term and the common difference, we can use the formula for the nth term of an arithmetic sequence: an = a1 + (n-1)d, where an is the nth term, a1 is the first term, n is the term number, and d is the common difference.

Using the given information, we can set up two equations:
14 = a1 + 4d (equation 1)
29 = a1 + 7d (equation 2)

Solving these equations simultaneously, we can find that a1 = 35 and d = 5.

To find the 12th term, we can substitute the values into the formula:
a12 = 35 + (12-1)5
a12 = 35 + 55
a12 = 90

Therefore, the 12th term of the sequence is 90.

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• 17.

### Dalam suatu barisan aritmatika, jika U3 + U7 = 56 dan U6 + U10 = 86 , maka suku ke-2 deret tersebut adalah ?

• A.

13

• B.

19

• C.

26

• D.

33

• E.

43

A. 13
Explanation
The given question is asking for the second term of the arithmetic sequence. To find the second term, we need to find the common difference (d) of the sequence. We can do this by subtracting U3 from U7 and U6 from U10. From the given equations, we have (U7 - U3) = 56 and (U10 - U6) = 86. Simplifying these equations, we get U4 = 56 and U4 = 86. Since these two equations contradict each other, it means that there is no unique common difference and therefore we cannot determine the second term of the sequence.

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• 18.

### Jumlah n suku pertama deret aritmetika dinyatakan dengan Sn = n2 + 2n. Beda dari deret itu adalah...

• A.

3

• B.

2

• C.

1

• D.

- 2

• E.

- 3

B. 2
Explanation
The given expression Sn = n^2 + 2n represents the sum of the first n terms of an arithmetic series. To find the common difference, we can subtract the expression for the sum of the first (n-1) terms from the expression for the sum of the first n terms. By doing so, we get (n^2 + 2n) - ((n-1)^2 + 2(n-1)) = 2. Therefore, the common difference of the arithmetic series is 2.

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• 19.

### Diketahui deret aritmatika 17, 20, 23, 26, ... Jumlah tiga puluh suku pertama deret tersebut adalah...

• A.

1815

• B.

2520

• C.

2310

• D.

2550

• E.

2810

A. 1815
Explanation
The given arithmetic series starts with 17 and has a common difference of 3. To find the sum of the first 30 terms, we can use the formula for the sum of an arithmetic series: Sn = (n/2)(2a + (n-1)d), where Sn is the sum of the first n terms, a is the first term, and d is the common difference. Plugging in the values, we get Sn = (30/2)(2(17) + (30-1)(3)) = 15(34 + 87) = 15(121) = 1815. Therefore, the correct answer is 1815.

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• 20.

### Diketahui barisan aritmetika  3, 8, 13, …. Suku keberapakah yang nilainya 198 ?

• A.

36

• B.

40

• C.

50

• D.

66

• E.

46

B. 40
Explanation
The given arithmetic sequence starts with 3 and has a common difference of 5. To find the nth term of an arithmetic sequence, we can use the formula an = a1 + (n-1)d, where an is the nth term, a1 is the first term, n is the position of the term, and d is the common difference. In this case, we need to find the position of the term that has a value of 198. Plugging in the values, we get 198 = 3 + (n-1)5. Simplifying the equation, we get n = 40. Therefore, the term with a value of 198 is the 40th term, which is the correct answer.

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• 21.

• A.

109

• B.

88

• C.

79

• D.

55

• E.

43

C. 79
• 22.

### Tentukan suku ke 10 dari barisan 1/8, 1/4, 1/2, ….

• A.

8

• B.

16

• C.

32

• D.

64

• E.

4

D. 64
Explanation
The given sequence is a geometric sequence with a common ratio of 2. To find the 10th term, we can use the formula for the nth term of a geometric sequence: an = a1 * r^(n-1), where a1 is the first term, r is the common ratio, and n is the term number. Plugging in the values, we get a10 = 1/8 * 2^(10-1) = 1/8 * 2^9 = 1/8 * 512 = 64. Therefore, the 10th term of the sequence is 64.

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• 23.

### .Diberikan sebuah deret geometri sebagai berikut.    3 + 6 + 12 + ..... Tentukan suku ke-5 dari deret tersebut!

• A.

48

• B.

24

• C.

96

• D.

108

• E.

60

A. 48
Explanation
The given series is a geometric series where each term is obtained by multiplying the previous term by a constant ratio. In this case, the ratio between each term is 2. Starting with the first term, which is 3, we can find the fifth term by multiplying the fourth term (12) by the ratio of 2. Therefore, the fifth term of the series is 24.

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• 24.

### Suku ke-n suatu deret geometri adalah 4-n. Maka jumlah tak hingga deret tersebut sama dengan

• A.

3

• B.

2

• C.

1

• D.

1/2

• E.

1/3

A. 3
Explanation
The given sequence represents the terms of a geometric series, where each term is obtained by subtracting the index of the term from 4. In this case, the first term (n=1) is 4-1=3. The sum of an infinite geometric series can be calculated using the formula S = a / (1 - r), where a is the first term and r is the common ratio. In this case, the common ratio is 1/2 (obtained by dividing each term by the previous term). Plugging in the values, S = 3 / (1 - 1/2) = 3 / (1/2) = 6. Therefore, the sum of the infinite geometric series is 6.

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• 25.

• A.

81

• B.

90

• C.

135

• D.

150

• E.

175

A. 81
• 26.

### Jika jumlah takhingga deret a + a0 + a-1 + a-2 + a-3 + … adalah 4a, maka nilai a adalah

• A.

4/3

• B.

2

• C.

3/2

• D.

3

• E.

4

A. 4/3
Explanation
The sum of an infinite geometric series is given by a / (1 - r), where a is the first term and r is the common ratio. In this case, the sum is 4a, so we can set up the equation 4a = a / (1 - r). Solving for r, we get r = 3/4. Since the common ratio is less than 1, the series converges. Therefore, the value of a is 4/3.

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• 27.

### Seorang anak menabung di suatu bank dengan selisih kenaikan tabungan antar bulan tetap.Pada bulan pertama sebesar Rp. 50.000,00, bulan kedua Rp.55.000,00, bulan ketiga Rp.60.000,00, dan seterusnya. Besar tabungan anak tersebut selama dua tahun adalah …

• A.

Rp. 1.315.000,00

• B.

Rp. 1.320.000,00

• C.

Rp. 2.040.000,00

• D.

Rp. 2.580.000,00

• E.

Rp. 2.640.000,00

E. Rp. 2.640.000,00
Explanation
The correct answer is Rp. 2.640.000,00. The question states that the savings increase by a fixed amount each month. The first month's savings is Rp. 50.000,00 and it increases by Rp. 5.000,00 each month. To find the total savings over two years, we need to calculate the sum of an arithmetic series. The formula for the sum of an arithmetic series is Sn = (n/2)(a + l), where Sn is the sum, n is the number of terms, a is the first term, and l is the last term. In this case, n is 24 (12 months in a year times 2 years), a is Rp. 50.000,00, and l is Rp. 2.640.000,00 (the last month's savings). Plugging these values into the formula, we get Sn = (24/2)(50.000 + 2.640.000) = 12(2.690.000) = Rp. 2.640.000,00.

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• 28.

### Sebuah mobil dibeli dengan hargaRp. 80.000.000,00. Setiap tahun nilai jualnya menjadi ¾ dari harga sebelumnya. Berapa nilai jual setelah dipakai 3 tahun ?

• A.

Rp. 20.000.000,00

• B.

Rp. 25.312.500,00

• C.

Rp. 33.750.000,00 Rp. 35.000.000,00

• D.

Rp. 45.000.000,00

• E.

Rp. 33.750.000,00

E. Rp. 33.750.000,00
Explanation
The value of the car decreases by 1/4 every year, which means it becomes 3/4 of the previous year's value. After 3 years, the value would be (3/4) * (3/4) * (3/4) * 80,000,000 = 33,750,000.

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• 29.

### Tentukan suku ke 10 dari barisan 1/8, 1/4, 1/2, ….

• A.

128

• B.

64

• C.

32

• D.

16

• E.

8

B. 64
Explanation
The given sequence is a geometric sequence with a common ratio of 2. Starting with the first term 1/8, we can find the 10th term by multiplying the first term by the common ratio raised to the power of 9 (10-1). So, the 10th term is (1/8) * (2^9) = 64.

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• 30.

### Diketahui barisan geometri, U2=14 dan U4=56, tentukan a dan rasionya?

• A.

7 dan 2

• B.

8 dan 3

• C.

9 dan 2

• D.

8 dan 2

• E.

7 dan 3

A. 7 dan 2
Explanation
Dalam barisan geometri, setiap suku (U) dapat ditemukan dengan rumus U = a * r^(n-1), dimana a adalah suku pertama, r adalah rasio, dan n adalah urutan suku tersebut. Dalam hal ini, kita memiliki U2 = 14 dan U4 = 56. Dengan menggantikan nilai n masing-masing dengan 2 dan 4, kita dapat membentuk dua persamaan: 14 = a * r^(2-1) dan 56 = a * r^(4-1). Dengan membagi persamaan kedua dengan persamaan pertama, kita dapat menghilangkan variabel a dan mendapatkan rasio r = 2. Dengan menggantikan nilai r = 2 ke dalam persamaan pertama, kita dapat menemukan nilai a = 7. Jadi, a = 7 dan rasio = 2.

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• 31.

### Diketahui suku pertama suatu deret geometri adalah 4 dengan suku ke-5 adalah 324. Tentukan rasio dari deret tersebut!

• A.

2 dan - 2

• B.

3 dan -3

• C.

4 dan - 4

• D.

1 dan -1

• E.

5 dan -5

B. 3 dan -3
Explanation
Dalam deret geometri, suku ke-n dapat ditemukan dengan mengalikan suku pertama dengan rasio pangkat (n-1). Dalam hal ini, suku pertama adalah 4 dan suku ke-5 adalah 324. Jadi, kita dapat menggunakan rumus ini untuk mencari rasio:
4 * (r^(5-1)) = 324
4 * r^4 = 324
r^4 = 81
r = 3 atau -3

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• 32.

### Jumlah tak hingga dari deret geometri   4+2+1+1/2+....

• A.

3,5

• B.

3

• C.

2,5

• D.

2

• E.

1

E. 1
Explanation
The given series is a geometric series with a common ratio of 1/2. As the common ratio is less than 1, the series converges to a finite value. To find the sum of the series, we can use the formula for the sum of an infinite geometric series, which is a/(1-r), where a is the first term and r is the common ratio. In this case, a = 4 and r = 1/2. Plugging these values into the formula, we get 4/(1-1/2) = 4/(1/2) = 4 * 2 = 8. Therefore, the sum of the given series is 8.

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• 33.

### Seutas tali dipotong menjadi 5 bagian sehingga potongan-potongan tali tersebut membentuk barisan geometri, jika panjang tali terpendek 6 cm dan potongan tali terpenjang 96 cm, maka panjang tali semula adalah  ....... cm?

• A.

205

• B.

186

• C.

175

• D.

160

• E.

145

B. 186
Explanation
The question states that a rope is cut into 5 pieces, forming a geometric sequence. The shortest piece is 6 cm and the longest piece is 96 cm. In a geometric sequence, each term is obtained by multiplying the previous term by a constant ratio. To find the length of the original rope, we need to find the common ratio. The common ratio can be found by dividing the longest piece by the shortest piece: 96 cm / 6 cm = 16. Therefore, the length of the original rope is the sum of the terms in the geometric sequence, which can be found using the formula: (first term * (1 - common ratio^n)) / (1 - common ratio). Plugging in the values, we get: (6 cm * (1 - 16^5)) / (1 - 16) = 186 cm.

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• 34.

### Diketahui suku ke 3 dan suku ke 5 suatu barisan geometri adalah 16 dan 64. Tentukan rasio dan jumlah sampai suku ke 6 ?

• A.

2/3

• B.

1/4

• C.

3

• D.

2

• E.

3/2

D. 2
Explanation
The given question provides information about the 3rd and 5th terms of a geometric sequence, which are 16 and 64 respectively. To find the ratio, we can divide the 5th term by the 3rd term, giving us 64/16 = 4. Therefore, the ratio of the geometric sequence is 4. To find the sum of the sequence up to the 6th term, we can use the formula S = a(1 - r^n) / (1 - r), where S is the sum, a is the first term, r is the ratio, and n is the number of terms. Plugging in the values, we get S = 16(1 - 4^6) / (1 - 4), which simplifies to S = 16(-4095) / (-3) = 5460. Therefore, the sum of the sequence up to the 6th term is 5460.

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• 35.

### Sebuah amoeba dapat membelah diri menjadi 2 setiap 6 menit. Pertanyaannya, berapakah jumlah amoeba setelah satu jam jika pada awalnya terdapat 2 amoeba?

• A.

1024

• B.

2048

• C.

4096

• D.

512

• E.

256

A. 1024
Explanation
The amoeba divides every 6 minutes, so in one hour (60 minutes), it will divide 10 times. Each time it divides, the number of amoebas doubles. Therefore, starting with 2 amoebas, after 10 divisions, there will be 2^10 = 1024 amoebas.

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• 36.

### Tentukan jumlah 6 suku pertama dari barisan 1,3,9,…

• A.

456

• B.

346

• C.

364

• D.

654

• E.

385

C. 364
Explanation
The given sequence is 1, 3, 9, ... In order to find the sum of the first 6 terms, we need to continue the pattern. The next term is 27, then 81, and so on. Adding the first 6 terms together, we get 1 + 3 + 9 + 27 + 81 + 243 = 364.

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• 37.

### Tentukan jumlah suku-suku deret geometri tak hingga dari 1 + 0,5 + 0,25 + 0,125 + ….

• A.

2

• B.

4

• C.

6

• D.

8

• E.

3

A. 2
Explanation
The given series is a geometric series with a common ratio of 0.5. In a geometric series, the sum of an infinite number of terms can be calculated using the formula S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio. In this case, a = 1 and r = 0.5. Plugging these values into the formula, we get S = 1 / (1 - 0.5) = 1 / 0.5 = 2. Therefore, the sum of the infinite number of terms in the series is 2.

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• 38.
E.
• 39.
• A.

85 m

• B.

80 m

• C.

75 m

• D.

70 m

• E.

60 m

D. 70 m
• 40.

### Diketahui a = 2 dan S∞ = 4, maka, nilai rasio adalah

• A.

2/5

• B.

1/4

• C.

1/2

• D.

2/3

• E.

3/5

C. 1/2
Explanation
The given question provides the values of a = 2 and S∞ = 4. The task is to determine the ratio. However, the question does not explicitly mention what the ratio is between. Without this information, it is not possible to generate an explanation for the given correct answer.

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• Mar 21, 2023
Quiz Edited by
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• Nov 07, 2016
Quiz Created by
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