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**Arithmetic Sequences (9.2)**

Common difference

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SAT Prep Quick poll!

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**Start with calculator review of sequences and partial sums**

There are three handy hand-outs: Sequences on the TI-84/84 Plus Partial sums on the TI 84/ 84 Plus Sequence mode on the TI 84/ 84 Plus Let’s see what sort of info they contain.

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**A sequence next Give the first five terms of the sequence for a1 = 2**

an+1 = an + 5 What is the pattern for the terms? What is a possible explicit formula?

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**A sequence next Give the first five terms of the sequence for**

an+1 = an + 5 What is the pattern for the terms? A common difference of 5.

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Arithmetic sequences If the pattern between terms in a sequence is a common difference, the sequence is arithmetic. The difference is d. Recursive: a1 an+1 = an + d So, d = an+1 - an

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Arithmetic sequences If the pattern between terms in a sequence is a common difference, the sequence is arithmetic. Explicit: an = a1 + (n-1) d (In other words, find the nth term by adding (n-1) d’s to the first term.)

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**Use it Give the first five terms for the sequence an = 2 + (n-1) 9**

Write the recursive formula.

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**Use it Give the first five terms for the sequence an = 2 + (n-1) 9**

2, 11, 20, 29, 38 Write the recursive formula. a1 = 2 an+1 = an + 9

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Use it an = 2 + (n-1) 9 Find the 10th term of the sequence.

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**Use it an = 2 + (n-1) 9 Find the 10th term of the sequence.**

= 2 + (9)9 = = 83

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Use it If the fourth term of a sequence is 5 and the ninth term is 20, find the sixth term. The key is to find the values for the first term and d. Start by writing the equations: 20 = a1 + (9 - 1) d 20 = a1 + 8d 5 = a1 + (4 - 1) d 5 = a1 + 3d

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Use it If the fourth term of a sequence is 5 and the ninth term is 20, find the sixth term. Next, use a system of equations to solve for d: 20 = a1 + 8d -(5 = a1 + 3d) 15 = 5d 3 = d

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Use it If the fourth term of a sequence is 5 and the ninth term is 20, find the sixth term. If d = 3, we can find the first term: 5 = a1 + (4 - 1) 3 5 = a1 + 9 a1 = -4

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Use it If the fourth term of a sequence is 5 and the ninth term is 20, find the sixth term. If d = 3, and a1 = -4 we can find the explicit equation: an = -4 + (n-1) 3 And the sixth term is a6 = -4 + (5)3 = 11.

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Partial sums Add the first 10 terms of our first sequence.

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**Partial sums Add the first 10 terms of our first sequence.**

How long did that take? Want a short cut?

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**Partial sums Add the first 10 terms of our first sequence.**

…+49 = 10(49). We want only half of this, so the sum is 5 (49), or 5(first term + last term), which is 245.

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**Partial sums In general, the partial sum for an arithmetic sequence is**

Sn = n/2 (a1 + an) or, if we substitute our explicit formula for an: Sn = n/2 (2a1 + (n - 1) d)

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**Partial sums In general, the partial sum for an arithmetic sequence is**

Sn = n/2 (a1 + an) Sn = n/2 (2a1 + (n - 1) d) Test it with our sequence: n = 10, a1 = 2, a10 = 47 S10 = 10/2(2 + 47) = 5(49) = 245

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Partial sums Find the sum of the first 100 positive integers.

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**Partial sums Find the sum of the first 100 positive integers.**

This problem was posed to Karl Friedrich Gauss ( ) in third grade, and he determined the pattern. S100 = 100/2(1+100) = 50(101) = 5050

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**Partial sums Find the sum of the first 50 positive even integers.**

How does this compare to the sum of the first 100 positive integers?

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**Partial sums Find the sum of the first 50 positive even integers.**

How does this compare to the sum of the first 100 positive integers? Let’s look at it in summation notation. S50 = 50/2( ) = 25(102) = 2550

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Warm Up: Section 2.11B Write a recursive routine for: (1). 6, 8, 10, 12,... (2). 1, 5, 9, 13,... Write an explicit formula for: (3). 10, 7, 4, 1,... (5).

Warm Up: Section 2.11B Write a recursive routine for: (1). 6, 8, 10, 12,... (2). 1, 5, 9, 13,... Write an explicit formula for: (3). 10, 7, 4, 1,... (5).

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