In Exercises 1-3, evaluate the arithmetic series. Using the formula an=a1+(n−1)d.

1. 1 + 2 + 3 +***+98+99+100

2. 500 + 490 + 480 +***+320+310+300

3. 2001 + 2002 + 2003 +***+4998+4999+5000

We will use the formula for the \(n\)-th term of an arithmetic series:

\[
a_n = a_1 + (n - 1) d
\]

where:
- \(a_n\) is the \(n\)-th term,
- \(a_1\) is the first term,
- \(d\) is the common difference, and
- \(n\) is the number of terms.

To find the sum of the arithmetic series, we also use the sum formula:

\[
S_n = \frac{n}{2} \cdot (a_1 + a_n)
\]

where:
- \(S_n\) is the sum of the first \(n\) terms,
- \(a_1\) is the first term,
- \(a_n\) is the last term,
- \(n\) is the number of terms.

### 1. \( 1 + 2 + 3 + \cdots + 100 \)

For this arithmetic series:
- First term, \(a_1 = 1\),
- Last term, \(a_n = 100\),
- Common difference, \(d = 1\).

Now, the number of terms \(n\) is 100. Using the sum formula:

\[
S_n = \frac{n}{2} \cdot (a_1 + a_n) = \frac{100}{2} \cdot (1 + 100) = 50 \cdot 101 = 5050
\]

So, the sum is **5050**.

### 2. \( 500 + 490 + 480 + \cdots + 300 \)

For this arithmetic series:
- First term, \(a_1 = 500\),
- Last term, \(a_n = 300\),
- Common difference, \(d = -10\).

Now, we find the number of terms \(n\). Using the formula for the \(n\)-th term:

\[
a_n = a_1 + (n - 1)d
\]
\[
300 = 500 + (n - 1)(-10)
\]
\[
300 = 500 - 10(n - 1)
\]
\[
300 - 500 = -10(n - 1)
\]
\[
-200 = -10(n - 1)
\]
\[
n - 1 = 20 \quad \Rightarrow \quad n = 21
\]

Now, we can calculate the sum:

\[
S_n = \frac{n}{2} \cdot (a_1 + a_n) = \frac{21}{2} \cdot (500 + 300) = 10.5 \cdot 800 = 8400
\]

So, the sum is **8400**.

### 3. \( 2001 + 2002 + 2003 + \cdots + 5000 \)

For this arithmetic series:
- First term, \(a_1 = 2001\),
- Last term, \(a_n = 5000\),
- Common difference, \(d = 1\).

Now, we find the number of terms \(n\). Using the formula for the \(n\)-th term:

\[
a_n = a_1 + (n - 1)d
\]
\[
5000 = 2001 + (n - 1)(1)
\]
\[
5000 = 2001 + n - 1
\]
\[
5000 = 2000 + n
\]
\[
n = 3000
\]

Now, we can calculate the sum:

\[
S_n = \frac{n}{2} \cdot (a_1 + a_n) = \frac{3000}{2} \cdot (2001 + 5000) = 1500 \cdot 7001 = 10501500
\]

So, the sum is **10,501,500**.

---

Would you like more details or have any questions? Here are some related problems you can try:

1. What is the sum of the series \(1 + 4 + 7 + \cdots + 100\)?
2. Evaluate the sum of \(100 + 90 + 80 + \cdots + 10\).
3. What is the sum of the first 50 terms of the series \(3 + 6 + 9 + \cdots\)?
4. Find the sum of \(1000 + 995 + 990 + \cdots + 500\).
5. Calculate the sum of \(10 + 15 + 20 + \cdots + 1000\).

**Tip:** For any arithmetic sequence, always check the common difference \(d\) before applying the formulas.

In Exercises 1-3, evaluate the arithmetic series. Using the formula an = a1 + (n−1)d.

1. 1 + 2 + 3 +***+98+99+100

2. 500 + 490 + 480 +***+320+310+300

3. 2001 + 2002 + 2003 +***+4998+4999+5000

Learn how to evaluate arithmetic series using the formulas for the nth term and the sum of an arithmetic sequence. This step-by-step guide solves three problems: summing 1 to 100, 500 to 300, and 2001 to 5000. Suitable for Grades 8-10 students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation