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Chapter 4: Problem 20

$$ 5+10+15+20+\cdots+300 $$

Short Answer

Expert verified
The sum of the arithmetic sequence from 5 to 300 with a common difference of 5 is 9302.5, with the sequence having 61 terms.

Step by step solution

01

Find the number of terms in the sequence (n)

Since we have an arithmetic sequence that starts from 5 and adds 5 each time, we can represent the nth term as: \[a_n = a_1 + (n - 1) * d\], where \(a_1 = 5\) is the first term, d is the common difference (which is 5), and n is the number of terms. We know that the last term is 300, so we have: \(a_n = 300\) \(5 + (n - 1) * 5 = 300\) Now we solve for n: \(5n - 5 = 300\) \(5n = 305\) \(n = 61\) So, there are 61 terms in the sequence.
02

Calculate the sum of the sequence

Now that we have the value of n, we can use the formula for the sum of an arithmetic sequence to find the sum: \(S_n = \frac{n}{2}(a_1 + a_n)\) \(S_{61} = \frac{61}{2} (5 + 300)\)
03

Evaluate the sum

Evaluate the sum by plugging in the values: \(S_{61} = \frac{61}{2} (305)\) \(S_{61} = 61 * \frac{305}{2}\) \(S_{61} = 61 * 152.5\) \(S_{61} = 9302.5\) So, the sum of the arithmetic sequence from 5 to 300 with a common difference of 5 is 9302.5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sum of Arithmetic Series
In an arithmetic series, we are presented with a sequence of numbers where the difference between consecutive terms is constant. The sum of an arithmetic series can be efficiently calculated using a specific formula.
The formula to discover the sum of an arithmetic sequence is:
  • \( S_n = \frac{n}{2} \times (a_1 + a_n) \)
where \( S_n \) is the sum of the sequence, \( n \) is the number of terms, \( a_1 \) is the first term, and \( a_n \) is the last term.
This formula essentially averages the first and last terms and multiplies by the number of terms, giving a precise outcome without the need of adding each individual term.
This makes it incredibly handy for long sequences, helping us avoid repetitive calculations.
Common Difference
The common difference is a cornerstone in an arithmetic sequence. It denotes the amount added (or subtracted if it is negative) to each term to get to the next one.
In the sequence given in the exercise, every term increases by 5, hence the common difference \( d = 5 \).
Know that:
  • The common difference value remains constant throughout the series.
  • It is a critical factor in determining both the next term in the sequence and the total number of terms.
To find a specific term in the arithmetic sequence, especially the nth term, you use:
  • \( a_n = a_1 + (n - 1) \times d \)
Understanding the common difference not only helps in predicting upcoming values but also plays a decisive role in calculating the series' sum effectively.
Number of Terms in Sequence
Calculating the number of terms in an arithmetic sequence is essential, especially when you want to find the series' sum.
The formula used to determine the total terms involves the first term, last term, and common difference:
  • \( n = \frac{a_n - a_1}{d} + 1 \)
In the original exercise, the sequence starts at 5 and ends at 300 with a common difference of 5. Substituting these into the formula will yield:
  • \( n = \frac{300 - 5}{5} + 1 = 61 \), confirming there are 61 terms.
Knowing the number of terms aspires towards several calculations and conditions:
  • Helps conveniently apply the sum formula for arithmetic series.
  • Determines the precise duration or length of the sequence.
Being able to calculate and understand the number of terms greatly simplifies many problems in arithmetic sequences and series.

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Most popular questions from this chapter

Use the well-ordering principle to prove that given any integer \(n \geq 1\), there exists an odd integer \(m\) and a nonnegative integer \(k\) such that \(n=2^{k} \cdot m\).

Write each of \(58-60\) as a single summation or product. $$ 2 \cdot \sum_{k=1}^{n}\left(3 k^{2}+4\right)+5 \cdot \sum_{k=1}^{n}\left(2 k^{2}-1\right) $$

Use strong mathematical induction to prove that for any integer \(n \geq 2\), if \(n\) is even, then any sum of \(n\) odd integers is even, and if \(n\) is odd, then any sum of \(n\) odd integers is odd.

The Archimedean property for the rational numbers states that for all rational numbers \(r\), there is an integer \(n\) such that \(n>r\). Prove this property.

Any product of two or more integers is a result of successive multiplications of two integers at a time. For instance, here are a few of the ways in which \(a_{1} a_{2} a_{3} a_{4}\) might be computed: \(\left(a_{1} a_{2}\right)\left(a_{3} a_{4}\right)\) or \(\left.\left(\left(a_{1} a_{2}\right) a_{3}\right) a_{4}\right)\) or \(a_{1}\left(\left(a_{2} a_{3}\right) a_{4}\right)\). Use strong mathematical induction to prove that any product of two or more odd integers is odd.
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