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

a

_{1}, a

_{2}, a

_{3}, a

_{4}, ... , a

_{n}, ...

is arithmetic if there is a number d such that

a

_{2}- a

_{1}= a

_{3}- a

_{2}= a

_{4}- a

_{3}= ... = 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:

a

_{2}= 1 + (2 - 1)4 = 1 + (1)(4) = 1 + 4 = 5

a

_{3}= 1 + (3 - 1)4 = 1 + (2)(4) = 1 + 8 = 9

a

_{4}= 1 + (4 - 1)4 = 1 + (3)(4) = 1 + 12 = 13

a

_{5}= 1 + (5 - 1)4 = 1 + (5)(4) = 1 + 20 = 21

a

_{2}- a

_{1}= a

_{3}- a

_{2}= a

_{4}- a

_{3}= ... = 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}= a

_{1}+ (n - 1) d

distributing d we get: a

_{n}= a

_{1}+ dn - d = dn + (a

_{1}- d) therefore let’s let c = (a

_{1}- d) so we have

a

_{n}= dn + c

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

_{1}- 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.

a

_{1}= 200, a

_{k+1}= a

_{k}- 20

a

_{2}= a

_{1+1}= a

_{1}- 20 = 200 - 20 = 180

a

_{3}= a

_{2+1}= a

_{2}- 20 = 180 - 20 = 160

a

_{4}= a

_{3+1}= a

_{3}- 20 = 160 - 20 = 140

a

_{5}= a

_{4+1}= a

_{4}- 20 = 140 - 20 = 120

so d = -20 and c = a

_{1}- 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.

a

_{1}= 3, a

_{2}= 13 , a

_{9}= ____

d = 13 - 3 = 10

c = a

_{1}- d = 3 - 10 = -7

so therefore a

_{n}= dn + c = 10n - 7

a

_{9}= 10n - 7 = 10 (9) - 7 = 90 - 7 = 83

Example 4: Find a formula for a

_{n}for the arithmetic sequence:

a

_{1}= 15, d = 4

c = a

_{1}- d = 15 - 4= 11

so therefore a

_{n}= dn + c = 4n +11

Example 5: Find a formula for an for the arithmetic sequence:

a

_{1}= -4, a

_{5}= 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 = a

_{1}- 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)(a

_{1}+ 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 = a

_{1}- d = - 6 - 4 = -10

so therefore a

_{n}= dn + c = 4n - 10

a

_{1}= -6 and a

_{50}= 4(50) - 10 = 200 - 10 = 190

S

_{n}= (n/2)(a

_{1}+ a

_{n}) = (n/2)(a

_{1}+ a

_{50}) = (50 / 2) (-6 + 190) = (25)(184) = 4600

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