FIRST ENCOUNTER WITH ARITHMETIC SEQUENCES (2)

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 u₁, u₂, u₃, …, u_n, … be any sequence. Based on the sequence we can generate another sequence s₁, s₂, s₃, s₄, …, s_n, … , where:

s₁ = u₁

s₂ = u₁ + u₂

s₃ = u₁ + u₂ + u₃

and so on.

In general, .

Moreover, s_n is called the nth partial sum of the sequence, i.e. the sum of the first n terms of the sequence u₁, u₂, u₃, …, u_n, … The sequence s₁, s₂, s₃, …, s_n, … is called the series generated by the sequence.

 

As an illustration, consider the arithmetic sequence 2, 5, 8, 11, 14, … In this sequence, u₁ = 2, u₂ = 5, u₃ = 8, and so forth. The nth term of the sequence is u_n = 2 + (n – 1)⋅3 = 3n – 1. Using this sequence we can generate the series s₁, s₂, s₃, …, s_n, … as follows.

s₁ = u₁ = 2

s₂ = u₁ + u₂ = 2 + 5 = 7

s₃ = u₁ + u₂ + u₃ = 2 + 5 + 8 = 15

s₄ = u₁ + u₂ + u₃ + u₄ = 2 + 5 + 8 + 11 = 26,

and so on.

The corresponding series is 2, 7, 15, 26, …, s_n, …

 

If u₁, u₂, u₃, …, 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 s₁₀₀. 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 u₁₀₀ by applying u_n = a + (n – 1)⋅b. Substituting the given values ​​into the u_n formula, we get: u₁₀₀ = 2 + (100 – 1)⋅3 = 299. Thus, we have to compute:

s₁₀₀ = 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 s₁₆. Substituting a = 135, u₁₆ = 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 nth partial sum is s_n = n² + 3n. Find the 30^th term of the sequence.

Answer

Note that u₃₀ = s₃₀ – s₂₉. Thus, u₃₀ can be obtained by subtracting s₂₉ from s₃₀.

s₃₀ = 30² + 3⋅30 = 990.

s₂₉ = 29² + 3⋅29 = 928

Consequently, the 30^th term of the sequence is u₃₀ = 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.