Progressions (Sequences)

In mathematics, progressions, or sequences, are ordered lists of numbers that follow a certain pattern. These patterns can be arithmetic or geometric in nature, and they are often used to represent real-world situations or solve mathematical problems.

In the case of the given exam question, we are presented with a sequence that appears to be an arithmetic progression. The given sequence is 2, 5, 8, 11, 14, … and we are asked to find the 20th term of this sequence.

To solve this problem, we first need to determine the common difference of the arithmetic progression. The common difference is the constant value that is added to each term to obtain the next term in the sequence. In this case, the difference between each consecutive term is 3, as 5 – 2 = 3, 8 – 5 = 3, and so on.

Now that we know the common difference is 3, we can find the 20th term of the progression. We can use the formula for the nth term of an arithmetic progression, which is given by:

an = a1 + (n – 1) * d

In this formula, an represents the nth term of the progression, a1 is the first term, n is the position of the term in the sequence, and d is the common difference.

Plugging in the values, we have:

a20 = 2 + (20 – 1) * 3

Simplifying this equation, we get:

a20 = 2 + 19 * 3
a20 = 2 + 57
a20 = 59

Therefore, the 20th term of the given arithmetic progression is 59. Thus, the correct answer to the exam question would be option C) 59.

In conclusion, progressions are a fascinating mathematical concept that allows us to understand and predict patterns in numbers. In this particular example, we used the knowledge of arithmetic progressions to find the 20th term of a given sequence. By understanding the pattern and applying the relevant formula, we can solve such problems with ease.
等差數列

在數學中，等差數列是按照一定模式排列的一系列數字。這些模式可以是等差或等比，通常用於表示現實世界的情況或解決數學問題。

在給定的考試題目中，我們遇到了一個看起來是等差數列的序列。給定的序列是2，5，8，11，14，…我們被要求找出這個序列的第20項。

要解決這個問題，我們首先需要確定等差數列的公差。公差是每個項目相加以得到下一個項目的固定值。在這種情況下，每兩個連續項之間的差值為3，例如5-2=3，8-5=3，以此類推。

既然我們知道公差是3，就可以找到等差數列的第20項。我們可以使用等差數列的第n項公式求解，該公式為：

an = a1 + (n – 1) * d

在這個公式中，an代表等差數列的第n項，a1是第一項，n是項在序列中的位置，d是公差。

代入數值，我們有：

a20 = 2 + (20 – 1) * 3

簡化方程，我們得到：

a20 = 2 + 19 * 3
a20 = 2 + 57
a20 = 59

因此，給定等差數列的第20項是59。因此，考試問題的正確答案為選項C）59。

總之，等差數列是一個迷人的數學概念，它使我們能夠理解和預測數字的模式。在這個特定的例子中，我們利用了等差數列的知識來找到給定序列的第20項。通過理解模式並應用相關的公式，我們可以輕松解決這樣的問題。