# JUSTIN AND JEREMY STARRING IN…. “SUMMER IS OVER AND BOY DID I MISS MATH!”

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JUSTIN AND JEREMY STARRING IN…. “SUMMER IS OVER AND BOY DID I MISS MATH!”

“WOW, J.K., SUMMER WAS GREAT AND ALL, BUT, BOY DID I MISS MATH.” “I KNOW WHAT YOU MEAN, J.M., ALL THOSE CRAZY SUMMER BEACH PARTIES JUST WEREN’T AS FUN AS A GOOD OLE’ EQUATION” “WHAT CAN WE DO??” “HEY, MAYBE COOL JANICE HAS AN IDEA!”

“Happy first day of school everyone!” “If you are looking for some math to do, I suggest doing some cool stuff with Arithmetic Sequences. Here is cool Ashley to start explaining…”

Well, the first thing we should consider is just what is an Arithmetic Sequence? Look at these license plates. The differences between each # and the previous # is always the same amount. What are the missing numbers? R2011 R20?? R2035

That’s right, the missing numbers are: R2011 R2019 R2027 R2035 Some of you may have done this in your head, others may have copied off of Keegan. Whatever. When numbers go up or down by the same amount, this is called an ARITHMETIC SEQUENCE.

What are some examples of arithmetic sequences in “real life”? Think of situations where you always add or subtract the same amount. Some examples are:  Buy one item for \$5; two items for \$10; three items for \$15; etc.  A basketball player shoots 1 basket and gets 2-points; 2 baskets equal 4 points; etc.  Every cake needs 3 eggs. For 1 cake you take away 3 eggs from the frig; for 2 cakes take away 6 eggs; etc.

What are some words we need to know in order to work with Arithmetic Sequences? Good question cool Cole. Let’s check it out……

Another thing to note is “n”. This letter just stands for what term you look at. In t 2 this stands for the 2 nd term. Here, t 2 has a value of 7.

So, look at the following sequence and then answer the questions: 9, 5, 1, -3, -7, -11 a) Is the sequence infinite or finite? b) What is the value of t 1 ? c) What is the value of t 5 ? d) What is the common difference, d ?

If you examine an arithmetic sequence, you can notice something about the number of times you need to add on your “d” to your first term: 6, 11, 16, 21 Here the t 1 = 6 and the d = 5 I can re-write the terms using t 1 and d : 6, 11, 16, 21 11 = 6 + 5,or, t 2 = t 1 + 1 X d 16 = 6 + 10,or, t 3 = t 1 + 2 X d 21 = 6 + 15,or, t 4 = t 1 + 3 X d We see that the number of times you multiply d is always one less than the term you are after. In other words: t n = t 1 + (n – 1) d

“Cool,” said cool Kailan. “If we want to find the value of any term in an arithmetic sequence, we can use: t n = t 1 + (n – 1) d I hope we get an example!”

t n = t 1 + (n – 1) d Solution: a) Use the formula: t n = t 1 + (n – 1) d t 20 = -3 + (20 – 1) (5) t 20 = -3 + (19)(5) = -3 + 95 = 92 b) Here they tell you the value (212) and you need to determine if it is t 30, t 31, t 32, etc. t n = t 1 + (n – 1) d 212 = -3 + (n – 1)(5) 212 + 3 = (n – 1)(5) 215 / 5 = n – 1  43 = n – 1  n = 44 212 is the 44 th term, or, t 44 = 212

“Life is good”, said Keegan. “It’s the first day of school and I – so far - understand my math.” “Don’t speak to soon”, warned Shaun. “Remember: Mr. Negrave hates kids. What will the next example be like?”

(Well, it isn’t too bad if you think about it…) Example: Solution: Use t n = t 1 + (n – 1) d twice! t 3 = t 1 + (3 – 1)d and t 8 = t 1 + (8 – 1)d 4 = t 1 + 2d and 34 = t 1 + 7d You have two equations with two things you don’t know. All you need to do is rearrange one equation then substitute into the remaining equation. Watch how Jill and Jimena handle it on the next page….

4 = t 1 + 2d and 34 = t 1 + 7d I’m going to rearrange the blue equation to get one variable by itself: 4 = t 1 + 2d 4 – 2d = t 1 Now, I substitute this in for “t 1 ” in the green equation. 34 = t 1 + 7d 34 = 4 – 2d + 7d 34 = 4 + 5d 30 = 5d d = 6 Now that I know d=6, I substitute it in order to get t 1 : t 1 = 4 – 2d = 4 – 2(6) = 4 – 12 = -8, or, t 1 = -8

“Amazing,” said Marian. “I totally get this and I didn’t even need to copy off of Ashley!” “I hope we get one to try!” said Daniel.

Ok, give this one a try and then we’ll check it…. A theatre has 25 seats in the second row and 65 seats in the seventh row. The last row has 209 seats. The number of seats in the rows produce an arithmetic sequence. a)How many seats are in the first row? b)How many rows of seats are in the theatre? c)Determine the “general term”, t n, for the sequence. (Note: All part c) means is for you to write an equation in the form tn = t 1 + (n – 1) d and substitute in the values for t 1 and d )

SOLUTION: a)Setup two equations and solve for t 1. t 2 = t 1 + (n – 1) d t 7 = t 1 + (n – 1) d 25 = t 1 + (2 – 1) d 65 = t 1 + (7 – 1) d 25 = t 1 + d 65 = t 1 + 6d t 1 = 25 – d  substitute  65 = 25 – d + 6d 65 = 25 + 5d 40 = 5d  d = 8 Therefore, t 1 = 25 – 8 = 17, or, there are 17 seats in row 1.

b) If the last row has 209 seats, use tn = t 1 + (n – 1) d in order to get “n” when the value of t n = 209 tn = t 1 + (n – 1) d 209 = 17 + (n – 1) 8 209 – 17 = (n – 1) 8 192 = (n – 1) 8 192 / 8 = n – 1 24 = n – 1 n = 25 There are 25 rows of seats in the theatre.

c) Here we simply substitute the values of t 1 and d into the general equation: tn = t 1 + (n – 1) d tn = 17 + (n – 1) 8 Is the general term for the sequence. Now, Tess and Meghan will let you know what questions to do!

Assignment: page 16 #1, 3, 4, 5b, 6b, 7, 11, 13, 16, 23, 24

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