Math Problem Statement

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

Solution

We will use the formula for the nn-th term of an arithmetic series:

an=a1+(n1)da_n = a_1 + (n - 1) d

where:

  • ana_n is the nn-th term,
  • a1a_1 is the first term,
  • dd is the common difference, and
  • nn is the number of terms.

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

Sn=n2(a1+an)S_n = \frac{n}{2} \cdot (a_1 + a_n)

where:

  • SnS_n is the sum of the first nn terms,
  • a1a_1 is the first term,
  • ana_n is the last term,
  • nn is the number of terms.

1. 1+2+3++1001 + 2 + 3 + \cdots + 100

For this arithmetic series:

  • First term, a1=1a_1 = 1,
  • Last term, an=100a_n = 100,
  • Common difference, d=1d = 1.

Now, the number of terms nn is 100. Using the sum formula:

Sn=n2(a1+an)=1002(1+100)=50101=5050S_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++300500 + 490 + 480 + \cdots + 300

For this arithmetic series:

  • First term, a1=500a_1 = 500,
  • Last term, an=300a_n = 300,
  • Common difference, d=10d = -10.

Now, we find the number of terms nn. Using the formula for the nn-th term:

an=a1+(n1)da_n = a_1 + (n - 1)d 300=500+(n1)(10)300 = 500 + (n - 1)(-10) 300=50010(n1)300 = 500 - 10(n - 1) 300500=10(n1)300 - 500 = -10(n - 1) 200=10(n1)-200 = -10(n - 1) n1=20n=21n - 1 = 20 \quad \Rightarrow \quad n = 21

Now, we can calculate the sum:

Sn=n2(a1+an)=212(500+300)=10.5800=8400S_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++50002001 + 2002 + 2003 + \cdots + 5000

For this arithmetic series:

  • First term, a1=2001a_1 = 2001,
  • Last term, an=5000a_n = 5000,
  • Common difference, d=1d = 1.

Now, we find the number of terms nn. Using the formula for the nn-th term:

an=a1+(n1)da_n = a_1 + (n - 1)d 5000=2001+(n1)(1)5000 = 2001 + (n - 1)(1) 5000=2001+n15000 = 2001 + n - 1 5000=2000+n5000 = 2000 + n n=3000n = 3000

Now, we can calculate the sum:

Sn=n2(a1+an)=30002(2001+5000)=15007001=10501500S_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++1001 + 4 + 7 + \cdots + 100?
  2. Evaluate the sum of 100+90+80++10100 + 90 + 80 + \cdots + 10.
  3. What is the sum of the first 50 terms of the series 3+6+9+3 + 6 + 9 + \cdots?
  4. Find the sum of 1000+995+990++5001000 + 995 + 990 + \cdots + 500.
  5. Calculate the sum of 10+15+20++100010 + 15 + 20 + \cdots + 1000.

Tip: For any arithmetic sequence, always check the common difference dd before applying the formulas.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Arithmetic Series
Algebra
Summation

Formulas

nth term formula: an = a1 + (n−1)d
Sum formula: Sn = (n/2) * (a1 + an)

Theorems

Arithmetic Series Theorem

Suitable Grade Level

Grades 8-10