Section 9.2 Notes Arithmetic Sequences and Series.

Section 9.2 Notes

Arithmetic Sequences and Series

Definitions

Sequence is a set of numbers arranged in a particular order.

Term is one of the set of numbers in a sequence.

Arithmetic Sequence is a sequence in which the difference between two consecutive numbers is constant.

Common difference is this constant difference.

Arithmetic Series is the indicated sum of the terms of an arithmetic sequence.

# of Term (n) 1 2 3 4 5

Term (an) 7 10 13 16 19

Suppose the following the 1st five terms of an arithmetic sequence are given

What would be the next term? 22

How did you find the next term? Added 3 to 19. 

You used the recursive formula of an arithmetic sequence.The recursive formula for an arithmetic sequence is   1 1, , 1 n na a a d n

If these numbers were graphed as ordered pairs (n, an) what type of function would it be?

Linear functionWhat does the common difference represent in this function?

SlopeWrite an equation for this arithmetic sequence in terms of an.

an = 3(n − 1) + 7

oran = 3n + 4

To find the value of any of the following in an arithmetic sequence:

a1 first term of the sequence

n which term of the sequence

d common difference of the sequence

an value of the nth term

Use

The explicit formula for an arithmetic sequence is

an = a1 + (n – 1)d

or

an = dn + c, where c is the zero term

C. To find “one” arithmetic mean between two numbers a and b use:

D. To find the sum of an arithmetic sequence (arithmetic series) use:

arithmetic mean = 2

a b

112

n n

nS a a

Or substitute a1 + (n − 1)d for an

1 1 12

n

nS a a n d

12 2 12

n

nS a n d

E. Solve the following arithmetic sequence or series problems.

1. Find n for the sequence for which

an = 633, a1 = 9, and d = 24

Steps:

a. Use: an = a1 + (n – 1)d

b. Substitute the given values and solve for the unknown value.

So 633 is the 27th term of the given sequence.

633 9 1 24 n

633 9 24 24 n

633 15 24 n

648 24 n27 n

2. Find the 58th term of the sequence:

10, 4, −2, …

Steps:

a. Find a1, n, and d.

  a1 = 10

n = 58

d = −6

b. Use: an = a1 + (n – 1)d

c. Substitute the given values and solve for the unknown value.

So the 58th term of the sequence is −332.

10 58 1 6 na

10 57 6 na

10 342 na

332na

An arithmetic mean of two numbers a and b is

simply their average

Numbers m1, m2, m3,… are called arithmetic means between a and b if a, m1, m2, m3,…, b forms an arithmetic sequence.

2

a b

Notice , , form an arithmetic sequence.2

a ba b

3. Form a sequence that has 5 arithmetic means between −11 and 19.

Steps:

a. The given sequence is

−11, m1, m2, m3, m4, m5, 19

b. a1 = −11, n = 7, an = 19

c. Use: an = a1 + (n – 1)d to solve for d.

So the arithmetic sequence is:

−11, −6, −1, 4, 9, 14, 19

19 11 7 1 d

19 11 6 d

30 6 d

5 d

4. Find n for a sequence for which a1 = 5,

d = 3, and Sn = 440.

1Use 2 1 to solve for .2n

nS a n d n

440 2 5 1 32

n

n

440 10 3 32

n

n

440 3 72

n

n

880 3 7 n n

2880 3 7 n n

20 3 7 880 n n

2880 3 7 n n

0 3 55 16 n n

n must be a positive integer so n = 16.

5. If a3 = 7 and a30 = 142, find a20.

Steps:

a. Write two explicit equations to make a system of equations to solve for a1 and d.

b. In the first equation use a30 = 142 and n = 30 to make the equation:

142 = a1 + (30 − 1)d or 142 = a1 + 29d.

c. In the second equation use a3 = 7 and n = 3 to make the equation:

7 = a1 + (3 − 1)d or 7 = a1 + 2d.

d. Solve for d in the system of equations.

1

1

142 29

7 2

a d

a d

1142 29 a d

17 2 a d

135 27 d5 d

e. Solve for a1.

17 2 5 a

17 10 a

13 a

f. Solve for a20.

20 3 20 1 5 a

20 3 19 5 a

20 3 95 a

20 92a

6. A lecture hall has 20 seats in the front row and two seats more in each following row than in the preceding one. If there are 15 rows, what is the seating capacity of the hall?

a1 = 20; d = 2; n = 15

15

1540 14 2

2S

12 12n

nS a n d

15

1568

2S

15 510 seatsS

Summation Notation

Summation notation is defined by

where i is called the index, n is the upper limit, and 1 is the lower limit.

1 2 31

...

n

i n ni

a a a a a S

1 2 3 11

...2

n

i n ni

na a a a a a a

Find the following sum.

14

1

3 1

n

n

1 3 1 1 4 a

14 3 14 1 43 a

14 4 432

329

To do a sum in a graphing calculator depends on the calculator.