Problem 36 - Algebra and Trigonometry

Question

A sequence is defined recursively. List the first five terms. $$a_{1}=3 ; \quad a_{n}=4-a_{n-1}$$

Step-by-Step Solution

Answer

The first five terms are 3, 1, 3, 1, 3.

Step 1 - Write down the first term

The first term of the sequence is given as \(a_1 = 3\).

Step 2 - Find the second term

To find the second term, use the recursive formula \(a_n = 4 - a_{n-1}\). Substituting \(a_1\): \(a_2 = 4 - a_1 = 4 - 3 = 1\).

Step 3 - Find the third term

Next, use the recursive formula for the third term: \(a_3 = 4 - a_2 = 4 - 1 = 3\).

Step 4 - Find the fourth term

Use the formula again for the fourth term: \(a_4 = 4 - a_3 = 4 - 3 = 1\).

Step 5 - Find the fifth term

Finally, use the formula to find the fifth term: \(a_5 = 4 - a_4 = 4 - 1 = 3\).

Step 6: Conclusion

The first five terms of the sequence are 3, 1, 3, 1, 3.

Key Concepts

Sequence Definition

In mathematics, a sequence is an ordered list of numbers following a particular pattern. Each number in the sequence is called a term. We often denote the terms of a sequence with a label, such as \( a_{n} \), where \( n \) represents the position of the term in the sequence.

First Term

The first term of a sequence is given explicitly and starts the sequence. In our exercise, the first term \( a_1 = 3 \) is provided. This value initializes the sequence and serves as a starting point for generating the subsequent terms.

Recursive Formula

A recursive formula defines each term of a sequence based on the preceding term(s). This means that you need to know the value of the previous term to find the next one. In the given problem, the formula is:
\( a_n = 4 - a_{n-1} \)
This tells us that to find any term \( a_n \), we subtract the previous term \( a_{n-1} \) from 4.

Pattern Identification

Identifying a pattern in a sequence can help predict future terms without computing each one individually. In this problem, we found the first five terms are 3, 1, 3, 1, and 3. Noticing this repetition, we see that the sequence alternates between 3 and 1. Recognizing such patterns can simplify generating terms beyond the initial few.