6.1 sequences and arithmetic sequences 1 a sequence is a set of numbers arranged in a definite...

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6.1 Sequences and Arithmetic Sequences

A sequence is a set of numbers arranged in a definite order. The following numbers form a sequence: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, . . .

In a sequence, the first term is denoted by a1, the second term is denoted by a2, the third term is denoted by a3, and so on. Using subscript notation, we can represent an infinite sequence by writing: a1, a2, a3, . . . , an, . . .

The term, an, represents the nth term, also called the general term.

The numbers in a sequence are called the terms of the sequence. If a sequence has an infinite number of terms, it is called an infinite sequence. A finite sequence has a first term and a last term. Used alone, the word, “sequence”, usually refers to an infinite sequence.

Calculating: a1 = (1 + 1)(1 - 1) = (2)(0) = 0

The general term of a sequence can be used to calculate terms of the sequence.Example 1. Find the first five terms of the sequence that has the general term: an = (n+1)(n-1)

a2 = (2 + 1)(2 - 1) = (3)(1) = 3

a3 = (3 + 1)(3 - 1) = (4)(2) = 8

a4 = (4 + 1)(4 - 1) = (5)(3) = 15

a5 = (5 + 1)(5 - 1) = (6)(4) = 24

Answer: In order, the first five terms are: 0, 3, 8, 15, and 24.

an = 62(n-1)

Your Turn Problem #1Find the first five terms for the sequence that has the general term:

Answer: In order, the first five terms are: 6, 12, 24, 48, and 96.

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An arithmetic sequence is a sequence in which the difference between a term and its preceding term is a constant.

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, . . .For the sequence above, the difference between a term and its preceding term is always 3. a2 - a1 = 5 - 2 = 3 , a3 - a2 = 8 - 5 = 3 , a4 - a3 = 11 - 8 = 3 ,a8 - a7 = 23 - 20 = 3 , a11 - a10 = 32 - 29 = 3 , …

The number, d, is called the common difference for the arithmetic sequence. For the arithmetic sequence above, the common difference is 3. For an arithmetic sequence, the general term is: an = a1 + (n-1)d, where a1 is the first term and d is the common difference.

Example 2. Find the general term for the arithmetic sequence; 7, -1, -9, -17, . . .The common difference, d = a2 - a1 = -1 - 7 = -8 .

(Note: we could have used the difference between any two consecutive terms, i.e.,d = a4 - a3 = -17 - (-9) = -8 )

Recall that in an arithmetic sequence, an = a1 + (n-1)d

So for the sequence above, an= 7 + (n-1)(-8)

= 7 - 8n + 8Answer: an = -8n + 15 Your Turn Problem #2Find the general term for the arithmetic sequence: 2, 5, 8, 11, …Answer: an = 3n - 1

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The first term and the common difference can be any real numbers.

Example 3. Find the general term of the arithmetic sequence1 2 1

, 1, 1 , 2 , 3, . . .3 3 3

1 2d = 1

3 3

21So a = (n 1)n 3 3

1 2 2n

3 3 3

2 1n

3 3

2 1Answer: a nn 3 3

Your Turn Problem #3

Find the general term for the arithmetic sequence:

1 1 1 12, 1 , , - , -1 , . . .

6 3 2 35 17

Answer: a nn 6 6

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For this sequence, d = 6 – 10 = -4

Example 4. Find the 40th term for the arithmetic sequence, 10, 6, 2, -2, . . .

For an arithmetic sequence, the formula for the general term can be used to find a desired term.

So, an = 10 + (n-1)(-4)

Since we want the 11th term, let n = 11.Thus, a11 = 10 + (40-1)(-4)

= 10 + (39)(-4)= 10 + (-156)= -146

Answer: The 11th term, a11 = -146

Answer: 26.7


Find the 17th term for the arithmetic sequence:

1.1, 2.7, 4.3, 5.9, . . .

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To find the sum of an arithmetic sequence, we need to know the first term, a1, the last term, an, and the number of terms, n.

1 nn

n a aS

2

The sum formula for the first n terms of an arithmetic sequence:

Example 5. Find the sum of the first 40 terms of the arithmetic sequence: 6, 10, 14, 18, . . .

For this sequence, d = 10 – 6 = 4

So, a40 = 6 + (40 – 1)(4)

Answer: The sum of the first 40 terms is 3360.

For this sum, n = 40

= 6 + (39)(4)

= 6 + 156

= 162

Using the sum formula with n = 40, we get

S402

162) 40(6

240(168)

26720

3360


Find the sum of the first 25 terms of the arithmetic sequence, 14, 7, 0, -7, . . .

Answer: S25 = -1750

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Example 6. Find the sum of the first 58 terms of the arithmetic sequence, 2, 31/2, 5, 61/2, . . .

For this sequence, d = 31/2 – 2 = 11/2 or 3/2

So, a58 = 2 + (58 – 1)(3/2)

For this sum, n = 58

= 2 + (57)(3/2)

= 2 + 171/2

= 175/2 or 871/2


S582

) 58(2 2

175

)29( 2

179

Answer: The sum of the first 58 terms, S58 = 2595 ½.

212595or

25191

Your Turn Problem #6Find the sum of the first 37 terms of the arithmetic sequence, 8, 52/5, 24/5, 1/5, . . .

37

7178 3Answer: S or 1435

5 5

7


For this sequence, a2 – a1 = 7, a3 – a2 = 7, a4 – a3 = 7. So, there is a common difference, d = 7, which means that we are, in fact, working with an arithmetic sequence. In order to use the sum formula for an arithmetic sequence, we must determine the number of terms, n, that are being added. We know that an = a1 + (n – 1)d.

Answer: The sum of the first 26 terms, S26 = 2431.

Example 7. Find 6 + 13 + 20 + 27 + . . . + 181

For this sum, we are not told that we are adding the terms of an arithmetic sequence. Therefore,we must first determine if this is, in fact, an arithmetic sequence. To do this, we must determine that there is a common difference between the terms.

So, 181 = 6 + (n – 1)7

26 = n 181 = 7n – 1


S26 2

)811 26(6

(187)13


Find 19 + 8 + (-3) + (-14) + . . . + (-102)

Answer: S12 = -498

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A sum of the consecutive terms of a sequence is called a series. An infinite series is the sum of the terms in an infinite sequence. A finite series is the sum of a finite number of consecutive terms from either a finite sequence or from an infinite sequence. A sum of consecutive terms from an arithmetic sequence is called an arithmetic series. The sum, Sn = a1 + a2 + a3 + . . . + an-1 + an , is a finite series because only a finite number of consecutive terms from a sequence are added. We can write the sum of a finite series in a more compact way using the Greek capital letter, sigma, S.

The letter, i, is called the index of summation (or just the index). Here, 1 (as part of the i = 1 under the S) is the lower limit of summation (or just the lower limit) and n (on top of the ) is the upper limit of summation (or just the upper limit). The index takes on all values from the lower limit to the upper limit, inclusive. Although we used the letter i as the index, other letters may be used for the index.

11 2 3 n-1 n

na a a a . . . a a ( the symbol means "is defined as")

i

i

Therefore, we have n terms. Example: )5(2)4(2)3(2)2(2)1(21

2

5

ii

30 terms5

108642

We read this symbol as, “the sum of ‘a’ sub-‘ i ’, as i goes from 1 to n.” It means to add all of the terms in a sequence starting from the first term and adding through the nth term.

na

1 ii

The sum, , has an lower limit of 1 and an upper limit of n.

na

1 ii

is called summation notation.

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Answer: The sum of the first 28 terms, S28 = 2240.

For this sequence, a1= 6(1) – 7 = 6 – 7 = -1 and a28= 6(28) – 7 = 168 – 7 =

161Using the sum formula with n = 28, we get

S28 28 -1 161

2

14(160) 2240

Example 8. Find

28

7) - (6i

1 i


Find

43

532 i

57

1 i 5246 1Answer: or 1049

5 5

So we can use the sum formula for arithmetic sequence. n = 28 (28 – 1 +1 = 28) and the difference d = 6.

Note that this sum of a finite series is actually the sum an arithmetic sequence. Replacing i with 1, 2, 3, . . .,

we get -1 + 5 + 11 + . . . + 161.

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We first note that the general term (3i + 1) defines an arithmetic sequence and that (63 – 12) + 1 = 52 terms from this sequence are to be added.

The first of these 52 terms to be added is a1 = 3(12) + 1 = 37.

The last of these 52 terms to be added is a52 = 3(63) + 1 = 190.

Because we are adding the terms of an arithmetic sequence, we can use the sum formula:

5902

2190 3752

2

a a52 S 521

52

Example 9. Find

63

i 12

3i 1

63

i 12

Answer: 3i 1 5902


Find 31

i 17

2i 5

Answer: -645

The End3-6-07

6.1 sequences and arithmetic sequences 1 a sequence is a set of numbers arranged in a definite...

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