Arithmetic Sequences (9.2) Common difference

SAT Prep Quick poll!

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.

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?

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.

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

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.)

Use it Give the first five terms for the sequence an = 2 + (n-1) 9 Write the recursive formula.

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

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

Use it an = 2 + (n-1) 9 Find the 10th term of the sequence. = 2 + (9)9 = 2 + 81 = 83

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

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

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

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.

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

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?

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

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)

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

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

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(1+100) = 50(101) = 5050

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?

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(2 + 100) = 25(102) = 2550