Problem 35 - Algebra and Trigonometry

Question

\(27-36\) . Determine the common difference, the fifth term, the nth term, and the 100 th term of the arithmetic sequence. $$ 2,2+s, 2+2 s, 2+3 s, \dots $$

Step-by-Step Solution

Answer

The common difference is \( s \), fifth term is \( 2+4s \), nth term is \( 2+(n-1)s \), and 100th term is \( 2+99s \).

Step 1: Identify the First Term

The first term of the sequence is given as \( a_1 = 2 \).

Step 2: Identify the Common Difference

In an arithmetic sequence, the common difference \( d \) is the difference between any two successive terms. From the given sequence: \ \( a_2 = 2 + s \) and \( a_1 = 2 \). \ Therefore, \( d = (2 + s) - 2 = s \).

Step 3: Find the Fifth Term

The fifth term \( a_5 \) can be found using the formula \( a_n = a_1 + (n-1)d \). \ For the fifth term, \( n = 5 \): \ \( a_5 = 2 + (5-1)s = 2 + 4s \).

Step 4: Express the nth Term

The nth term of an arithmetic sequence is expressed by the formula \( a_n = a_1 + (n-1)d \). \ Substituting \( a_1 = 2 \) and \( d = s \), we have: \ \( a_n = 2 + (n-1)s \).

Step 5: Compute the 100th Term

Use the nth-term formula to find the 100th term by setting \( n = 100 \). \ \( a_{100} = 2 + (100-1)s = 2 + 99s \).

Key Concepts

Common Difference

The common difference in an arithmetic sequence is a vital concept. It represents the consistent amount added to each term to get to the next term. In simpler words, it is what makes the sequence tick forward steadily. To find the common difference, subtract the first term from the second term.

Let's see it in action with our sequence: the first term is 2, and the second term is \(2 + s\). To find the common difference (denoted as \(d\)), we calculate:

• \(d = (2 + s) - 2 = s\)

Understanding this helps us to get a grip on how each term is connected and moves along in the sequence.

Nth Term Formula

The nth term formula is a powerful tool in arithmetic sequences. It allows you to find any term in the sequence without having to list all the terms up to that point, which saves a lot of time!

The formula for the nth term \(a_n\) is given by:

• \(a_n = a_1 + (n-1) \cdot d\)

This formula requires three things:

• The first term \(a_1\), which in our case is 2.
• The common difference \(d\), which is \(s\).
• The term number \(n\), which is the position in the sequence you're interested in.

With the given sequence, our nth term turns into:

• \(a_n = 2 + (n-1)s\)

Grasping this formula means you can jump straight to any term in the sequence.

Fifth Term

To find the fifth term in the sequence, you can use the nth term formula we talked about earlier. By substituting \(n = 5\) into the equation, you will find:

• \(a_5 = a_1 + (5-1) \cdot d\)

For our sequence, this translates to:

• \(a_5 = 2 + 4s\)

The fifth term is simply 2 plus four times the common difference \(s\). This direct application of the formula helps understand how each position in the sequence is structured.

100th Term

Similarly, to find the 100th term of the sequence, the nth term formula comes to the rescue again. By plugging in \(n = 100\), you can directly determine:

• \(a_{100} = a_1 + (100-1) \cdot d\)

For this sequence:

• \(a_{100} = 2 + 99s\)

This term gives a sense of how far the sequence has progressed by the time you reach the 100th position. The formula simplifies the process, rendering the need to write out every term unnecessary. It is a big time-saver in tasks requiring large terms.