Arithmetic Sequences (9.2)

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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
• Lets 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
• an1 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
• a1 2 2, 7, 12, 17, 22
• an1 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
• an1 an d So, d an1 - 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) ds 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
• an1 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.
• a10 2 (10 - 1) 9
• 2 (9)9
• 2 81
• 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.
• 2 7 12 17 22 27 32 37 42 47
• How long did that take? Want a short cut?

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Partial sums

• Add the first 10 terms of our first sequence.
• 2 7 12 17 22 27 32 37 42 47
• 47 42 37 32 27 22 17 12 7 2
• 49494949 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
  (1777-1855) in third grade, and he determined the
  pattern.
• S100 100/2(1100) 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? Lets look at it in summation
  notation.
• S50 50/2(2 100) 25(102) 2550