Part 1 of this article introduces the concept of arithmetic sequence. Like sequences in general, arithmetic sequences can generate what we call arithmetic series. In brief, for every term of an arithmetic sequence there corresponds a partial sum; the sequence whose terms are the partial sums is called the * arithmetic series*.

**The Partial Sums and Series**

Let u_{1}, u_{2}, u_{3}, …, u_{n}, … be any sequence. Based on the sequence we can generate another sequence s_{1}, s_{2}, s_{3}, s_{4}, …, s_{n}, … , where:

s_{1} = u_{1}

s_{2} = u_{1} + u_{2}

s_{3} = u_{1} + u_{2} + u_{3}

and so on.

In general, .

Moreover, s_{n} is called the *n*th * partial sum* of the sequence, i.e. the sum of the first

*n*terms of the sequence u

_{1}, u

_{2}, u

_{3}, …, u

_{n}, … The sequence s

_{1}, s

_{2}, s

_{3}, …, s

_{n}, … is called the

*generated by the sequence.*

**series**

As an illustration, consider the arithmetic sequence 2, 5, 8, 11, 14, … In this sequence, u_{1} = 2, u_{2} = 5, u_{3} = 8, and so forth. The *n*th term of the sequence is u_{n} = 2 + (n – 1)⋅3 = 3n – 1. Using this sequence we can generate the series s_{1}, s_{2}, s_{3}, …, s_{n}, … as follows.

s_{1} = u_{1} = **2**

s_{2} = u_{1} + u_{2} = 2 + 5 = **7**

s_{3} = u_{1} + u_{2} + u_{3} = 2 + 5 + 8 = **15**

s_{4} = u_{1} + u_{2} + u_{3} + u_{4} = 2 + 5 + 8 + 11 = **26**,

and so on.

The corresponding series is 2, 7, 15, 26, …, s_{n}, …

If u_{1}, u_{2}, u_{3}, …, u_{n}, … is an arithmetic sequence then:

…………………………………………………………………………………………………………………………………………………………………………… (1)

…………………………………………………………………………………………………………………………………………………………… (2)

**Example 1**

Find the sum of the first 100 terms of the sequence 2, 5, 8, 11, 14, …

**Answer**

The sequence is an arithmetic sequence with the first term 2 and the common difference 3. Thus, a = 2 and b = 3. We have to find s_{100}. Substitute a = 2, b = 3, and n = 100 into (2). This results in:

Accordingly, the sum of the first 100 terms of the sequence is 15050.

The problem in Example 1 can also be solved using (1). First calculate u_{100} by applying u_{n} = a + (n – 1)⋅b. Substituting the given values into the u_{n} formula, we get: u_{100} = 2 + (100 – 1)⋅3 = 299. Thus, we have to compute:

s_{100} = 2 + 5 + 8 + 11 + 14 + … + 299.

By (1), we have .

**Example 2**

Compute 135 + 131 + 127 + 123 + … + 75.

**Answer**

The sequence 135, 131, 127, 123, … is an arithmetic sequence with the first term a = 135 and the common difference b = -4. To find the sum, we need to know first which term is 75. In other words, we have to find the value of n such that u_{n} = 75. To answer this, use the formula u_{n} = a + (n – 1).b. Substituting a, b, and u_{n} into the formula, we get:

75 = 135 + (n – 1)⋅(-4)

75 = 139 – 4n

From this we get n = 16.

Thus, what is required in this example is s_{16}. Substituting a = 135, u_{16} = 75, and n = 16 into (1), we obtain:

.

Therefore, 135 + 131 + 127 + 123 + … + 75 = 1680.

**Example 3**

Given an arithmetic sequence with the property that its *n*th partial sum is s_{n} = n^{2} + 3n. Find the 30^{th} term of the sequence.

**Answer**

Note that u_{30} = s_{30} – s_{29}. Thus, u_{30} can be obtained by subtracting s_{29} from s_{30}.

s_{30} = 30^{2} + 3⋅30 = 990.

s_{29} = 29^{2} + 3⋅29 = 928

Consequently, the 30^{th} term of the sequence is u_{30} = 990 – 928 = 62.

**Example 4**

Find n if 2 + 4 + 6 + … + 2n = 600.

**Answer**

In this case, a = 2 and b = 2. Substituting these values into (2), we get:

By the definition of a sequence, . Thus, n = -25 is not the solution to the equation. The remaining possibility is n = 24.