Problem 45 - A Graphical Approach to Precalculus with Limits

Question

Find \(a_{1}\) and \(d\) for each arithmetic sequence. $$S_{20}=1090, a_{20}=102$$

Step-by-Step Solution

Answer

\(a_1 = 7\) and \(d = 5\).

Step 1: Understanding the Problem

We are given an arithmetic sequence with the sum of the first 20 terms, \(S_{20} = 1090\), and the 20th term, \(a_{20} = 102\). Our task is to find the first term \(a_1\) and the common difference \(d\) of this sequence.

Step 2: Using the Formula for the nth term

The formula for the nth term in an arithmetic sequence is \(a_n = a_1 + (n-1)d\). Using this formula for \(a_{20} = 102\), we can write: \[ 102 = a_1 + 19d \]

Step 3: Using the Formula for the Sum of n Terms

The formula for the sum of the first n terms in an arithmetic sequence is \(S_n = \frac{n}{2}(a_1 + a_n)\). Using this formula for \(S_{20} = 1090\), we have:\[ 1090 = \frac{20}{2} (a_1 + 102) = 10(a_1 + 102) \]

Step 4: Solving the Equation for the Sum

We simplify the equation for the sum:\[ 1090 = 10(a_1 + 102) \] Divide both sides by 10:\[ 109 = a_1 + 102 \] Subtract 102 from both sides:\[ a_1 = 7 \]

Step 5: Substituting to Find the Common Difference

Now that we have \(a_1 = 7\), substitute into the equation for the 20th term:\[ 102 = 7 + 19d \] Subtract 7 from both sides:\[ 95 = 19d \] Divide both sides by 19:\[ d = 5 \]

Step 6: Conclusion

We found that the first term \(a_1 = 7\) and the common difference \(d = 5\).

Key Concepts

Sum of Arithmetic Sequence

The sum of an arithmetic sequence, often referred to with the symbol \(S_n\), is a critical concept when working with sequences of numbers where each term after the first is found by adding a fixed, constant number. This constant is called the extit{common difference}. To find the sum of the first \(n\) terms of an arithmetic sequence, we use the formula:\[ S_n = \frac{n}{2} (a_1 + a_n) \]Here's how it works:

• \(n\) is the total number of terms you want to add up.
• \(a_1\) is the first term of the sequence.
• \(a_n\) is the last term of the sequence you're summing up to.

This formula efficiently gives the total sum by calculating the average of the first and last terms, then multiplying by the number of terms. In our example, the first 20 terms add up to 1090. Understanding how to apply this formula allows you to tackle various problems involving arithmetic sequences with confidence.

Common Difference

The common difference in an arithmetic sequence is the consistent value that you add to one term to get to the next. This core element, denoted by the letter \(d\), is what defines the sequence's uniformity and pattern.You can find the common difference with a simple subtraction: take any two consecutive terms from the sequence and subtract the first from the second. Mathematically, if you know the nth and the (n+1)th term: \[ d = a_{n+1} - a_n \]In cases where more information is provided, such as a specific term several positions along the sequence, you can rearrange and use given equations to solve for \(d\). For instance, in our sample exercise, after you find the first term, you can plug it into the formula for any specific term to find \(d\). Without understanding this concept, determining the sequence's progression becomes impossible.

Nth Term Formula

The \(n\)th term of an arithmetic sequence is the formula used to calculate any term in the sequence without listing all the preceding terms. This formula is essential because it offers a direct means to find any desired term positionally within the sequence quickly.The formula is:\[ a_n = a_1 + (n-1)d \]Where:

• \(a_n\) is the term you're solving for.
• \(a_1\) is the first term in the sequence.
• \(d\) is the common difference.
• \(n\) is the term number you are trying to find.

To use this formula effectively, you start with the first term and repeatedly add the common difference. For instance, if you need the 20th term (as shown in our exercise), this formula allows you to ascertain it without manually adding the common difference for each step. Its straightforward application is ideal for expanding your ability to solve related math problems efficiently.