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Arithmetic Sequences

A sequence in which each term after the first is obtained by adding a fixed number to the previous term is an arithmetic sequence (or arithmetic progression). The fixed number that is added is the common difference. The sequence

5,9,13,17,21,

is an arithmetic sequence since each term after the first is obtained by adding 4 to the previous term.

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Arithmetic Sequences

That is,4

4

9 5

13 9

17 13

21 1 47 ,

4

and so on. The common difference is 4.

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Arithmetic Sequences

If the common difference of an arithmetic sequence is d, then by the definition of an arithmetic sequence,

for every positive integer n in the domain of the sequence.

1 ,n nd a a Common difference d

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Example 1 FINDING THE COMMON DIFFERENCE

Find the common difference, d, for the arithmetic sequence

9, 7, 5, 3, 1, Solution We find d by choosing any two adjacent terms and subtracting the first from the second. Choosing – 7 and – 5 gives

5 ( 7) .2d Choosing – 9 and – 7 would give the same result.

7 ( 9) .2d

Be careful when subtracting a

negative number.

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Example 2 FINDING TERMS GIVEN a1 AND d

Find the first five terms for each arithmetic sequence.

Solution

a. The first term is 7, and the common difference is – 3.

1 7a Start with a1 = 7.

2 ( 37 4)a Add d = – 3.

3 ( 34 1)a Add – 3.

4 31 2( )a Add – 3.

5 ( 32 5)a Add – 3.

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Example 2 FINDING TERMS GIVEN a1 AND d

Find the first five terms for each arithmetic sequence.

Solution

b. a1 = – 12, d = 5

1 12a Start with a1.

2 12 75a Add d = 5.

3 57 2a

4 2 35a

5 853a

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Look At It This WayIf a1 is the first term of an arithmetic sequence and d is the common difference, then the terms of the sequence are given by

1 1a a

12a da

13 2 1 2a a d a d d a d

34 1 1 32a a d a d d a d

15 4a a d

16 ,5da a

and, by this pattern,

1 ( 1 .)n n da a

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nth Term of an Arithmetic Sequence

In an arithmetic sequence with first term a1 and common difference d, the nth term, is given by

1 ( 1) .na a n d

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Example 3FINDING TERMS OF AN ARITHMETIC SEQUENCE

Find a13 and an for the arithmetic sequence – 3, 1, 5, 9, …Solution

Here a1 = – 3 and d = 1 – (– 3) = 4. To find a13 substitute 13 for n in the formula for the nth term.

1 1 ( 1)a a n d

113 (13 1)a a d n = 13

Work inside parentheses

first.

13 3 (12)4a Let a1 = – 3, d = 4.

13 3 48a Simplify.

13 45a

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Example 3FINDING TERMS OF AN ARITHMETIC SEQUENCE

Find a13 and an for the arithmetic sequence – 3, 1, 5, 9, …Solution

Find an by substituting values for a1 and d in the formula for an.

( 1 4)3na n Let a1 = – 3, d = 4.

Simplify.

3 4 4na n Distributive property

4 7na n

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Arithmetic Series

The total amount of interest paid is given by the sum of the terms of this sequence.Now we develop a formula to find this sum without adding all 30 numbersdirectly. Since the sequence is arithmetic, we can write the sum of the first n terms as

1 1 1 12 ( 1) .nS a a d a d a n d

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Sum of the First n Terms of an Arithmetic Sequence

If an arithmetic sequence has first term and common difference d, then the sum of the first n terms is given by

1( )2n n

nS a a or 1[2 ( 1) ].

2n

nS a n d

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Example 7 USING THE SUM FORMULAS

Evaluate S12 for the arithmetic sequence – 9, – 5, – 1, 3, 7, ….

a.

Solution

We want the sum of the first 12 terms. Using a1 = – 9, n = 12, and d = 4 in the second formula,

1[2 ( 1) ],2n

nS a n d

gives 12

12[2( 9) 11(4)] 156.

2S

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Example 7 USING THE SUM FORMULAS

b.

Solution

Use a formula for Sn to evaluate the sum of the first 60 positive integers.

The first 60 positive integers form the arithmetic sequence 1, 2, 3, 4, …, 60. Thus, n = 60, a1 = 1, and a60 = 60, so we use the first formula in the preceding box to find the sum.

1( )2n n

nS a a

60

60(1 60) 1830

2S