Problem 76
Question
Solve each problem. Supports on a Slide\(\quad\) A slide with a uniform slope is to be built on a level piece of land. There are to be 20 equally spaced supports, with the longest support 15 meters long and the shortest 2 meters long. Find the total length of all the supports.
Step-by-Step Solution
Verified Answer
Total length of supports is 170 meters.
1Step 1: Understanding the Problem
We need to find the total length of 20 supports on a slide with equally spaced lengths. The shortest support is 2 meters, and the longest is 15 meters.
2Step 2: Identify Patterns
Since the supports are equally spaced in length, they form an arithmetic sequence. The first term of this sequence, \(a_1\), is 2 meters, and the last term, \(a_{20}\), is 15 meters.
3Step 3: Calculate Common Difference
The formula for the nth term of an arithmetic sequence is \(a_n = a_1 + (n-1)d\). Plug in the values: \(15 = 2 + 19d\). Solving for \(d\), we get \(d = \frac{13}{19}\).
4Step 4: Find the Total Length Using the Sum Formula
The formula for the sum of the first \(n\) terms of an arithmetic sequence is \(S_n = \frac{n}{2}(a_1 + a_n)\). Substituting the known values: \(S_{20} = \frac{20}{2}(2 + 15) = 10 \times 17 = 170\).
5Step 5: Verification
Verify by finding the middle term arithmetic sum, confirming the total length matches expectations based on calculations.
Key Concepts
Understanding Arithmetic SeriesCalculating the Sum of SeriesRole of Common DifferenceProblem Solving with Arithmetic Series
Understanding Arithmetic Series
An arithmetic series consists of the sum of the terms in an arithmetic sequence. In simple terms, when the sequence of numbers has a common difference between each term, their sum forms an arithmetic series. This is essential when you're adding multiple terms that are regularly increasing or decreasing, like the supports in our slide problem. In the example given, each support is a term in the series, and we add all these terms to find the total length of the supports. When you sum up all the terms in a sequence that progresses with a constant amount, you harness the concept of arithmetic series. This method of addition helps streamline calculations by using formulas.
Calculating the Sum of Series
To find the sum of an arithmetic series efficiently, use the formula: \[ S_n = \frac{n}{2} (a_1 + a_n) \] This formula means you add the first term \(a_1\) and the last term \(a_n\) together, then multiply by the number of terms \(n\) divided by 2.
This approach ingeniously pairs terms from the start and end of the series, maximizing efficiency and ensuring accuracy. For our slide problem, substituting 2 and 15 as the first and last terms respectively, with 20 supports, gives a total length of 170 meters. By focusing on the beginning and end values plus how many terms there are, we're simplifying what could be tedious addition.
This approach ingeniously pairs terms from the start and end of the series, maximizing efficiency and ensuring accuracy. For our slide problem, substituting 2 and 15 as the first and last terms respectively, with 20 supports, gives a total length of 170 meters. By focusing on the beginning and end values plus how many terms there are, we're simplifying what could be tedious addition.
Role of Common Difference
The common difference, denoted as \(d\), plays a pivotal role in identifying patterns within an arithmetic sequence. It's the uniform amount added or subtracted between consecutive terms. In the supports example, moving from the shortest to the longest support in consistent steps can be captured through this common difference. To calculate \(d\), use the formula for any term in the series: \[ a_n = a_1 + (n-1)d \] By first solving \(15 = 2 + 19d\), we find \(d = \frac{13}{19}\). The common difference is crucial because it ensures the sequence progresses uniformly, influencing both the final term and the sum of the series.
Problem Solving with Arithmetic Series
Solving problems involving arithmetic series requires understanding the sequence and leveraging the formulae effectively. It's about identifying the key elements: the first term, the last term, the number of terms, and the common difference. Having a structured approach is vital. This involves:
Successfully identifying these parts and using the formulas saves time and increases accuracy. Confirming results by checking calculations reinforces the reliability of the answer, as shown when verifying that the total length of the supports indeed reached the expected 170 meters. Practicing these steps aids in mastering similar mathematical challenges.
- Recognizing the progression is arithmetic
- Calculating the common difference
- Applying the sum formula
Successfully identifying these parts and using the formulas saves time and increases accuracy. Confirming results by checking calculations reinforces the reliability of the answer, as shown when verifying that the total length of the supports indeed reached the expected 170 meters. Practicing these steps aids in mastering similar mathematical challenges.
Other exercises in this chapter
Problem 75
Prove each statement for positive integers \(n\) and \(r,\) with \(r \leq n\) (Hint: Use the definitions of permutations and combinations.) $$P(n, n-1)=P(n, n)$
View solution Problem 75
Use summation notation to write each series. Start the index at \(i=1\). $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{9}$$
View solution Problem 76
Prove each statement for positive integers \(n\) and \(r,\) with \(r \leq n\) (Hint: Use the definitions of permutations and combinations.) $$P(n, 1)=n$$
View solution Problem 76
Use summation notation to write each series. Start the index at \(i=1\). $$-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\frac{1}{81}-\dots-\frac{1}{2187}$$
View solution