Download Section 11.2 Arithmetic Sequences

Section 11.2
Arithmetic Sequences
Objectives:
•Recognizing arithmetic sequences by
their formulas
•Finding the first term, the common
difference, and find partial sums.
Definition of Arithmetic Sequence
An arithmetic sequence is in which the difference
between each term and the preceding term is
always constant.
For example: 2,4,6,8,… is an arithmetic sequence
because the difference between each term is 2.
1,4,9,16,25,… is not arithmetic because the
difference between each term is not the same.
Ex.1 Are the following sequences arithmetic? If
so, what is the common difference.
a) 14, 10, 6, 2, -2, -6, -10,…
Yes, the sequence is arithmetic because the
difference between each term is constant.
The common difference is -4
b) 3, 5, 8, 12, 17, …
No, the sequence is not arithmetic because the
difference between each term in not the
same.
Formula for nth term of arithmetic
sequence:
The formula for the nth term of an arithmetic
sequence is
an = a1 + (n – 1)d
Ex 2. If an is an arithmetic sequence with a1= 3
and a2 = 4.5 then
a) Find the first 5 terms of the sequence
The first 5 terms are 3, 4.5, 6, 7.5, and 9
b) Find the common difference, d.
d = 1.5
Ex3. Find the first six terms and the 300th term
of the arithmetic sequence 13, 7,…
Since a1=13 and a2=7 then the common
difference is 7 - 13 = -6.
So the nth term is
an = a1 + (n – 1)d
an = 13 – 6(n – 1)
• From that, we find the first six terms:
13, 7, 1, –5, –11, –17, . . .
• The 300th term is:
a300 = 13 – 6(299)
= –1781
Ex 4. The 11th term of an arithmetic sequence is
52, and the 19th term is 92. Find the 1000th term.
To find the nth term of the sequence,
we need to find a and d in the formula
an = a + (n – 1)d
From the formula, we get:
a11 = a + (11 – 1)d = a + 10d
a19 = a + (19 – 1)d = a + 18d
• Since a11 = 52 and a19 = 92, we get
the two equations:
52  a  10d

92  a  18d
– Solving this for a and d, we get: a = 2, d = 5
– The nth term is: an = 2 + 5(n – 1)
– The 1000th term is: a1000 = 2 + 5(999) = 4997
Partial Sums of Arithmetic Sequences
For the arithmetic sequence an = a + (n – 1)d,
the nth partial sum is given by either of these
formulas.
n
1. Sn  2a1  (n  1)d 
2
 a1  an 
2. Sn  n 

2


Ex 5. Find the sum of the first 40 terms of
the arithmetic sequence 3, 7, 11, 15, . . .
– Here, a = 3 and d = 4.
– Using Formula 1 for the partial sum of
an arithmetic sequence, we get:
S40 = (40/2) [2(3) + (40 – 1)4]
= 20(6 + 156)
= 3240
Ex 6. Find the sum of the first 50 odd numbers.
– The odd numbers form an arithmetic
sequence with a = 1 and d = 2.
– The nth term is:
an = 1 + 2(n – 1) = 2n – 1
– So, the 50th odd number is:
a50 = 2(50) – 1 = 99
Substituting in Formula 2 for the partial sum of
an arithmetic sequence,
we get:
 a  a50 
 1  99 
S50  50 
 50 


 2 
 2 
 50  50
 2500
Ex 7. An amphitheater has 50 rows of seats with 30
seats in the first row, 32 in the second, 34 in the
third, and so on. Find the total number of seats.
• The numbers of seats in the rows form an
arithmetic sequence with a = 30 and d = 2.
– Since there are 50 rows, the total number
of seats is the sum
S50  502 2(30)  49(2)
 3950
– The amphitheater has 3950 seats.
Ex 8 How many terms of the arithmetic
sequence 5, 7, 9, . . . must be added to get 572?
We are asked to find n when Sn = 572.
• Substituting a = 5, d = 2, and Sn = 572
in Formula 1, we get:
n
572 
2  5   n  1 2
2
572  5n  n  n  1
0  n 2  4n  572
0   n  22  n  26 
– This gives n = 22 or n = –26.
– However, since n is the number of terms
in this partial sum, we must have n = 22.
HW p837 1-49 odd