Problem 43 - Precalculus: Mathematical for Calculus

Question

Find the partial sum \(S_{n}\) of the arithmetic sequence that satisfies the given conditions. $$a_{1}=55, d=12, n=10$$

Step-by-Step Solution

Answer

The partial sum \( S_{10} \) is 1090.

Step 1: Identify the Formula

To find the partial sum of an arithmetic sequence, we use the formula: \[ S_n = \frac{n}{2} (a_1 + a_n) \] where \( n \) is the number of terms, \( a_1 \) is the first term, and \( a_n \) is the nth term.

Step 2: Calculate the Nth Term

The nth term \( a_n \) of an arithmetic sequence can be calculated using the formula: \[ a_n = a_1 + (n-1) \cdot d \] Substituting the given values, we find \( a_{10} = 55 + (10-1) \cdot 12 = 55 + 108 = 163 \).

Step 3: Substitute Values into the Sum Formula

With \( a_{10} = 163 \), substitute into the sum formula: \[ S_{10} = \frac{10}{2} (55 + 163) = 5 \cdot 218 \].

Step 4: Compute the Partial Sum

Calculate the product to find \( S_{10} \): \[ S_{10} = 5 \cdot 218 = 1090 \].

Key Concepts

Understanding Arithmetic Sequences

In mathematics, an arithmetic sequence, or arithmetic progression, is a sequence of numbers in which the difference between consecutive terms is constant. Understanding this constant difference is crucial as it defines the entire sequence.
For example, if you start with the number 55 and add 12 to get the next term, you are forming an arithmetic sequence. Here, 55 is the first term, and 12 is the common difference, represented by \(d\).

• The general form of an arithmetic sequence can be expressed as: \(a, a+d, a+2d, a+3d, \ldots\)
• This shows that each term after the first is obtained by adding \(d\) to the previous term.

The arithmetic sequence continues in this pattern indefinitely, forming a predictable and easy-to-track list of numbers.

What is a Partial Sum?

A partial sum, in the context of arithmetic sequences, is the sum of the first \(n\) terms of the sequence. It's a way of finding the total of a specified number of terms without having to individually add each one.
In our exercise, we are asked to find the partial sum of the first 10 terms using the initial term \(a_1 = 55\) and a common difference of \(d = 12\).

• Finding a partial sum allows you to see how much these specific parts of the sequence add up to.
• It is particularly useful in practical situations where you need to sum a portion of the sequence rather than the whole.

Partial sums provide valuable insights into the behavior and properties of sequences, especially when analyzing patterns over a set period or term count.

The Sum Formula for Arithmetic Sequences

To efficiently calculate the sum of terms in an arithmetic sequence, you can use the sum formula:\[ S_n = \frac{n}{2} (a_1 + a_n) \]
Here, \(n\) is the number of terms for which you want the sum, \(a_1\) is the first term, and \(a_n\) is the last term to be included in the sum.
This formula is derived from pairing terms in the sequence such that each pair sums to the same value, simplifying the addition process.

• The sum formula allows a quick evaluation without needing to add each number individually.
• It's especially helpful when dealing with large sequences, saving both time and effort.

In our case, substituting the values \(n=10\), \(a_1=55\), and \(a_{10}=163\) into the formula gives the total sum for these terms.

Calculating the Nth Term

To find the nth term in an arithmetic sequence, use the following calculation formula: \(a_n = a_1 + (n-1) \cdot d\).
For our example, this involves substituting \(a_1 = 55\), \(d = 12\), and \(n = 10\) into the formula, leading to \(a_{10} = 55 + 9 \cdot 12\). This simplifies to \(a_{10} = 163\).
The nth term shows us the actual member of the sequence at position \(n\) and is essential for finding the partial sum using the sum formula.

• Understanding how to calculate the nth term helps to locate specific terms within a sequence.
• It is crucial for not only sum calculations but any mathematical analysis requiring pinpointing an exact sequence position.

With the nth term calculation in hand, solving complex arithmetic progression problems becomes more manageable.