Happy Monday!!! (Groundhog Day) 2-02-15 Sequences Day 4&5 Happy Monday!!! (Groundhog Day) 2-02-15

Warm-Up Find the sum: 𝑘=4 10 𝑘 2 −4 𝑘=0 3 2𝑘+1 𝑘=4 10 𝑘 2 −4 𝑘=0 3 2𝑘+1 Write the sum with out sigma notation: 𝑘=1 7 𝑘 −1

Arithmetic Sequence One of the simplest ways to generate a sequence is to start with a number 𝑎 and add to it a fixed constant 𝑑 over and over again. An arithmetic sequence is a sequence in the form 𝑎, 𝑎+𝑑, 𝑎+2𝑑, 𝑎+3𝑑, 𝑎+4𝑑… The number 𝑎 is the first term, and 𝑑 is the common difference of the sequence.

𝒂 𝒏 =𝒂+ 𝒏−𝟏 𝒅 Arithmetic Sequence The nth term of an Arithmetic Sequence is given by: 𝒂 𝒏 =𝒂+ 𝒏−𝟏 𝒅 Again 𝒅 is the common difference because any two consecutive terms of an arithmetic sequence differ by 𝒅.

Example: Arithmetic Sequence EX1: if 𝑎=2 and 𝑑=3 the arithmetic sequence is 2, 2+3 , 2+3 +3 ,… or 2,5,8,… Since 𝑎=2 and common difference, 𝑑=3 The nth term is 𝒂 𝒏 =𝟐+𝟑 𝒏−𝟏

Ex2 Given the sequence: 9,4,−1,−6,−11 Find the nth term of the sequence: First find common difference: 𝑑= 𝑎 2 − 𝑎 1 →4−9=−5 2) Use the formula: 𝑎 𝑛 =𝑎+𝑑 𝑛−1 𝑎 𝑛 =9−5 𝑛−1

You try Find the first 6 terms and the 300th term of the arithmetic sequence given the first two terms are 13, 7.

Solutions First term 𝑎=13. Common Difference is 𝑑=7−13=−6 𝑠𝑜: 𝑎 𝑛 =13−6 𝑛−1 𝐹𝑖𝑟𝑠𝑡 6 𝑡𝑒𝑟𝑚𝑠: 13, 7, 1,−5,−11,−17 300th term : 𝑎 300 =13−6 299 =−1781

Finding terms The 11th term of an arithmetic sequence is 52 and the 19th terms is 92. Find the 1000th term. To find the nth term of this sequence we need to find 𝑎 and 𝑑 𝑎 𝑛 =𝑎+ 𝑛−1 𝑑 From this formula we get: 𝑎 11 =𝑎+ 11−1 𝑑=𝑎+10𝑑 𝑎 19 =𝑎+ 19−1 =𝑎+18𝑑

Continued Since 𝑎 11 =52 and 𝑎 19 =92 we have system to solve 52=𝑎+10𝑑 92=𝑎+18𝑑 Using elimination: −40=−8𝑑 so d=5 Substituting 𝑑=5 we find 52=𝑎+10 5 → 52=𝑎+50→𝑎=2 So to find the nth term: 𝒂 𝒏 =𝟐+ 𝒏−𝟏 𝟓

1000th term So now all we need to do is use the formula we found. 𝑎 𝑛 =2+5 𝑛−1 𝑎 1000 =2+5 1000−1 =2+5 999 𝑎 1000 =4997

You try Is the sequence arithmetic? if yes find the common difference. 2, 13, 24, 35… 25,1,−23, −47… log 5 , log 7 , log 8… 4 , 16 , 36 , 64 … The 12th term of an arithmetic sequence is 36, and the sixth term is 18. Find the 20th .

Answers 𝒂 𝟐𝟎 =𝟑+𝟑 𝟏𝟗 =𝟔𝟎 1) Yes , d=11 2) Yes, d= -24 3) No 5) 𝑎 𝑛 =𝑎+𝑑 𝑛−1 𝑎 12 =36=𝑎+11𝑑 𝑎 6 =18 =𝑎+5𝑑 −18=−6𝑑 →𝒅=𝟑 36=𝑎+ 11∗3 →36−33=𝟑=𝒂 𝒂 𝟐𝟎 =𝟑+𝟑 𝟏𝟗 =𝟔𝟎

Homework Page 831- 832 #1- 6 and 7-16 even

Day 5: Arithmetic Sequences & geometric sequences 2-3-15

Partial sums of Arithmetic Sequences You individually have 3 minutes to find the sum of the numbers 1 to 100. 𝑘=1 100 𝑘 Ready? Go! What sum did you get?

Gauss When the famous mathematician C.F. Gauss was a school boy his professor gave him this same problem expecting to keep all the students busy for a while. But Gauss answered the questions almost immediately. His idea was this: since we are adding numbers produced according to a fixed pattern, there must also be a pattern (or formula) for finding the sum.

Gauss He started by writing the numbers 1 to 10 and below them the same number in reverse order. Writing S for the sum and adding corresponding terms. 𝑆=1+2+3+…+98+99+100 𝑆=100+99+98+…+3+2+1 2𝑆=101+101+101+…+101+101+101 Or 2𝑆=100 101 =5050

What does this mean? We can apply this idea to any arithmetic sequence Gauss and others derived 2 formulas To find the sum of any arithmetic sequence 𝑆 𝑛 = 𝑛 2 2𝑎+ 𝑛−1 𝑑 These are the same just represented differently 2) 𝑆 𝑛 =𝑛 𝑎+ 𝑎 𝑛 2

Find the partial sum Find the sum of the first 40 terms of the arithmetic sequence 3,7,11,15 1st : Find 𝑎 and 𝑑 𝒏=𝟒𝟎 𝐢𝐬 𝐠𝐢𝐯𝐞𝐧 𝑎=3 𝑑=7−3=4 Plug in to formula 𝑆 40 = 40 2 2 3 + 40−1 ∗4 = 𝟐𝟎 𝟔+𝟏𝟓𝟔 =𝟑𝟐𝟒𝟎

Using the other formula: EX Find the partial sum of the first 50 odd numbers. 𝑎=1 the difference between any two consecutive odd terms is 2, 𝑑=2 . So the 50th odd number is 1+ 50−1 ∗2=99 𝑆 50 =50 1+99 2 = 50∗50=2500

Word problem 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.

Solution 𝑎=30, 𝑑=2 𝑎𝑛𝑑 , 𝑛=50 𝑔𝑖𝑣𝑒𝑛 𝑆 50 = 50 2 2 30 +49 2 =𝟑𝟗𝟓𝟎 𝒔𝒆𝒂𝒕𝒔

Good afternoon Happy Wednesday! Let’s take out our homework! Warm-Up: 1) Find the sum of the first 150 even terms.

Questions on homework?

Finding the number of terms in a partial sum How many terms of the arithmetic sequence 5,7,9… must added to get 572. This means to solve for n. 𝑎=5, 𝑑=2 1st: 𝑆 𝑛 =572= 𝑛 2 2 5 + 𝑛−1 ∗2 2nd simplify and distribute: 572=𝑛 5 +𝑛 𝑛−1 572= 𝑛 2 +4𝑛 → 𝒏 𝟐 +𝟒𝒏 −𝟓𝟕𝟐 So 𝑛−22 𝑛+26 ⇒𝑛=22 𝑜𝑟 𝑛=−26 𝑛 is number of rows so 𝒏=𝟐𝟐

Classwork Practice Page 831: 36-40 Lets work 36 together

Practice Arithmetic Sequences Given the series 5,8,11,14… Find the sum of the first 12 terms 2) Given the series 21,15,9,3… Find the sum of the first 20 terms

Practice Sequences 3) Given 𝑎 18 =3362 and 𝑎 38 =7362 Find the equation for the nth term 4) How many terms of the arithmetic sequence -3, 2, 7, ... must be added together for the sum of the series to be 116?

Homework Page 832: #23-28, 29, 31,

Proof/ explanation Partial sum of arithmetic sequence. Well we can apply what Gauss did to any arithmetic sequence. 𝑆 𝑛 =𝑎+ 𝑎+𝑑 +…+ 𝑎+ 𝑛−2 𝑑 +[𝑎+ 𝑛−1 𝑑] 𝑆 𝑛 = 𝑎+ 𝑛−1 𝑑 + 𝑎+ 𝑛−2 𝑑 +…+ 𝑎+𝑑 +𝑎 Sum this up to get 2 𝑆 𝑛 = 2a+ n−1 d + 2a+ n−1 d +…+ 2a+ n−1 d + 2a+ n−1 d 2 𝑆 𝑛 =𝑛 2𝑎+ 𝑛−1 𝑑 Or get 𝑆 𝑛 by itself by dividing by 2 𝑺 𝒏 = 𝒏 𝟐 [𝟐𝒂+ 𝒏−𝟏 𝒅]