Latihan Soal Deret Dan Baris Aritmatika

Explanation
The given formula Sn = 2+4n represents the sum of the first n terms of an arithmetic series. To find the 9th term, we substitute n = 9 into the formula. Sn = 2+4(9) = 2+36 = 38. Therefore, the 9th term of the arithmetic series is 38.

Explanation
The monthly profit for the merchant increases by Rp. 18,000. So, to find the total profit until the 12th month, 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 of the series, n is the number of terms, a is the first term, and l is the last term. In this case, n = 12, a = Rp. 46,000, and l = Rp. 46,000 + (12-1) * Rp. 18,000 = Rp. 46,000 + Rp. 198,000 = Rp. 244,000. Plugging in these values into the formula, we get Sn = (12/2)(Rp. 46,000 + Rp. 244,000) = 6 * Rp. 290,000 = Rp. 1,740,000. Therefore, the correct answer is A. Rp. 1,740,000.

Explanation
The given information states that the 5th term of the arithmetic sequence is 22 and the 12th term is 57. To find the common difference, we subtract the 5th term from the 12th term, which gives us 57 - 22 = 35. Now, we can use the common difference to find the 15th term. We add 35 to the 12th term to get 57 + 35 = 92. Therefore, the 15th term of the arithmetic sequence is 92, which corresponds to option C. 72.

Explanation
The given question is asking for the sum of the first thirty terms of an arithmetic series. We are given that the 6th term is 17 and the 10th term is 33. From this information, we can find the common difference, which is 4. Using the formula for the sum of an arithmetic series, Sn = (n/2)(2a + (n-1)d), where a is the first term, n is the number of terms, and d is the common difference, we can substitute the values to find the sum. Plugging in a = 17, d = 4, and n = 30, we get Sn = (30/2)(2(17) + (30-1)(4)) = 15(34 + 29(4)) = 15(34 + 116) = 15(150) = 1,650. Therefore, the correct answer is A. 1,650.

Explanation
The formula for the sum of the first n terms of an arithmetic series is Sn = (n/2)(2a + (n-1)b), where a is the first term and b is the common difference. In this question, the sum of the first n terms is given as Sn, so we can rewrite the formula as Sn = (n/2)(2a + (n-1)b). Simplifying this equation, we get Sn = n(a + b(n-1)/2). Comparing this equation with the options, we can see that option C, 2a + b(2n+1), matches the given equation. Thus, option C is the correct answer.