9.2 Arithmetic Sequences and Partial Sums

9.2 Arithmetic Sequences and Partial Sums

Arithmetic Sequences - a sequence whose consecutive terms have a common difference

A. Definition of Arithmetic Sequences:
A sequence is arithmetic if the differences between consecutive terms are the same. So, the sequence

a₁, a₂, a₃, a₄, ... , a_n, ...

is arithmetic if there is a number d such that

a₂ - a₁ = a₃ - a₂ = a₄ - a₃ = ... = d.

The number d is the common difference of the arithmetic sequence.

Example 1: Given the sequence a_n = 1 + (n - 1)4, find the first five terms of the sequence an the common difference:
a₂ = 1 + (2 - 1)4 = 1 + (1)(4) = 1 + 4 = 5
a₃ = 1 + (3 - 1)4 = 1 + (2)(4) = 1 + 8 = 9
a₄ = 1 + (4 - 1)4 = 1 + (3)(4) = 1 + 12 = 13
a₅ = 1 + (5 - 1)4 = 1 + (5)(4) = 1 + 20 = 21

a₂ - a₁ = a₃ - a₂ = a₄ - a₃ = ... = d.

9 - 5 = 13 - 9 = 21 - 13 = 4 = d

B. The nth term of an Arithmetic Sequence: last year you learned it as

a_n = a₁ + (n - 1) d
distributing d we get: a_n = a₁ + dn - d = dn + (a₁ - d) therefore let’s let c = (a₁ - d) so we have

a_n = dn + c

where d is the common difference between consecutive terms of the sequence and c = a₁ - d

Example 2: write the first five terms of the arithmetic sequence, find the common difference and write the nth term of the sequence as a function of n.

a₁ = 200, a_k+1 = a_k - 20
a₂ = a₁₊₁ = a₁ - 20 = 200 - 20 = 180
a₃ = a₂₊₁ = a₂ - 20 = 180 - 20 = 160
a₄ = a₃₊₁ = a₃ - 20 = 160 - 20 = 140
a₅ = a₄₊₁ = a₄ - 20 = 140 - 20 = 120
so d = -20 and c = a₁ - d = 200 - -20 = 220
therefore a_n = dn + c = -20n + 220

Example 3: The first two terms of the arithmetic sequences are given. Find the missing term.
a₁ = 3, a₂ = 13 , a₉ = ____

d = 13 - 3 = 10
c = a₁ - d = 3 - 10 = -7
so therefore a_n = dn + c = 10n - 7

a₉ = 10n - 7 = 10 (9) - 7 = 90 - 7 = 83

Example 4: Find a formula for a_n for the arithmetic sequence:
a₁ = 15, d = 4

c = a₁ - d = 15 - 4= 11
so therefore a_n = dn + c = 4n +11

Example 5: Find a formula for an for the arithmetic sequence:
a₁ = -4, a₅ = 16

You may have noticed by now that a_n = dn + c is a linear function so therefore d is the slope of the line so using this concept of the slope formula ▵y/▵x We have:

(16 - -4) / (5 - 1) = (20)/ (4) = 5 = d

c = a₁ - d = - 4 - 5 = -9
so therefore a_n = dn + c = 5n - 9

C. The Sum of a Finite Arithmetic Sequence

Carl Friedrich Gauss (1777 - 1855) was ten years old, his teacher asked him to add all the integers from 1 to 100. Gauss was able to answer the teacher within a few moments, the teacher was amazed. Here is how he did it:

1 + 2 + 3 + 4 + ... + 97 + 98 + 99 + 100

He took the 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101 so taking 100 / 2 = 50, then 50 * 101 = 5050.

The Sum of a Finite Arithmetic Sequence:
The sum of a finite arithmetic sequence with n terms is

S_n = (n/2)(a₁ + a_n)

Example 6: find the indicated nth partial sum of the arithmetic sequence:

-6, -2, 2, 6, ... , n = 50

the common difference is -2 - -6 = 2 - -2 = 6 - 2 = 4
c = a₁ - d = - 6 - 4 = -10
so therefore a_n = dn + c = 4n - 10

a₁ = -6 and a₅₀ = 4(50) - 10 = 200 - 10 = 190
S_n = (n/2)(a₁ + a_n) = (n/2)(a₁ + a₅₀) = (50 / 2) (-6 + 190) = (25)(184) = 4600

9.2 Homework #53 page 635; # 9 - 19 odd, 35 - 47 odd, 59 - 79 odd