# Section 11.2 Arithmetic Sequences - PowerPoint PPT Presentation

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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.

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Section 11.2 Arithmetic Sequences

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## 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.

• 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

• 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:

• 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.

### 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:

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

• 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:

• This gives n = 22 or n = –26.

• However, since n is the number of terms in this partial sum, we must have n = 22.