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IB Studies Level Mathematics Arithmetic Sequences and Series

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The Family Block of Chocolate Imagine if I gave you a family block of chocolate which was made up of 100 small squares every day. Imagine if I gave you a family block of chocolate which was made up of 100 small squares every day. However as each day passed I started to eat a few pieces before I gave it to you. However as each day passed I started to eat a few pieces before I gave it to you. On the second day I eat three pieces so you get 97 pieces. On the second day I eat three pieces so you get 97 pieces. On the third day I eat six pieces, on the fourth I eat nine pieces! On the third day I eat six pieces, on the fourth I eat nine pieces! So the amount of chocolate you get every day is … So the amount of chocolate you get every day is … How much chocolate would you get altogether? How much chocolate would you get altogether?

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Arithmetic Series This series of numbers ( …) is called an arithmetic series. This series of numbers ( …) is called an arithmetic series. We can easily solve this problem with the right Information and tools! We can easily solve this problem with the right Information and tools!

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Arithmetic Sequences and series Have a look at the following sequences: Have a look at the following sequences: An Arithmetic series is a series of numbers in which each term is obtained from the previous term by adding or subtracting a constant. An Arithmetic series is a series of numbers in which each term is obtained from the previous term by adding or subtracting a constant. The constant we add or subtract each time is called the common difference, d In our chocolate example this was -3. The first term is called a (here it was 100). The constant we add or subtract each time is called the common difference, d In our chocolate example this was -3. The first term is called a (here it was 100). The letter n is used to denote the number of terms The letter n is used to denote the number of terms

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Algebraically T 1 1 st term T 2 2 nd term T 3 3 rd term T 4 4 th term T n nth term aa+ da+2da+3d? So for any term, n T n = a +(n-1) d

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Example Find the 20 th term of the sequence Find the 20 th term of the sequence 5,8,11,14,… 5,8,11,14,… Here a = 5 and d= 3 Here a = 5 and d= 3 So T n = 5+ (20-1)3 So T n = 5+ (20-1)3 T n = 62 T n = 62

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So what about our chocolate problem? We need to know how to sum an arithmetic sequence in order to solve our chocolate problem. We need to know how to sum an arithmetic sequence in order to solve our chocolate problem. Heres a neat proof to show you the formula. Heres a neat proof to show you the formula.

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The Sum of an Arithmetic Series- S n Let the last term of an Arithmetic series be l. Let the last term of an Arithmetic series be l. S n = a + a+d + a+2d + a+3d + … + l-d + l Eqn (1) S n = a + a+d + a+2d + a+3d + … + l-d + l Eqn (1) Now re-writing this backwards! Yes backwards! Now re-writing this backwards! Yes backwards! S n = l + l-d + l-2d + l-3d + …. +a+d + a Eqn (2) S n = l + l-d + l-2d + l-3d + …. +a+d + a Eqn (2) We are now going to add the two equations together- can you see why? What cancels out? We are now going to add the two equations together- can you see why? What cancels out?

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The Sum of an Arithmetic Series- S n So 2S n = lots of (a+l) but how many if there are n terms? So 2S n = lots of (a+l) but how many if there are n terms? Yes there are n lots of (a+l) Yes there are n lots of (a+l) This gives 2S n = n(a+l) This gives 2S n = n(a+l) But what if we dont know the last term? But what if we dont know the last term? S n = n/2 (a+l)

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The Sum of an Arithmetic Series- S n We can use l = a+ (n-1)d because l is the nth term of the series so substituting We can use l = a+ (n-1)d because l is the nth term of the series so substituting S n = n/2 ((a + a+(n-1)d)) S n = n/2 ((a + a+(n-1)d)) Which gives Which gives S n = n/2 (2a + (n-1)d) S n = n/2 (2a + (n-1)d) Heres a nice applet Heres a nice applet Heres a nice applet Heres a nice applet

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Example using the summation formula Find the sum of the first 22 terms of the arithmetic series …. Find the sum of the first 22 terms of the arithmetic series …. Using S n = n/2 (2a + (n-1)d) Using S n = n/2 (2a + (n-1)d) S n = 11 ( (-2)) S n = 11 ( (-2)) S n = -330 S n = -330

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How many pieces of chocolate? Our series for the choc0late problem looks like this: …..+ Our series for the choc0late problem looks like this: …..+ Here there are 34 terms n=34 since 100/3 Here there are 34 terms n=34 since 100/3 So S 34 = 34/2( (-3)) So S 34 = 34/2( (-3)) S 34 = 1717 S 34 = 1717 So you will eat 1717 pieces of chocolate after 34 days. So you will eat 1717 pieces of chocolate after 34 days. How many pieces will you eat on the last day? How many pieces will you eat on the last day?

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A nice summary of APs Here is a quick summary if you need it. Here is a quick summary if you need it. Here is a quick summary if you need it. Here is a quick summary if you need it. Answers to sheet Answers to sheet

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