Times Eleven and Arithmetic Progression

Mentor: How many facts are there in the first and the last uncolored rows together?

Student: 1+10=11

Mentor: What about the second row and the row before the last (the ninth) together? I am doing pairs of rows this way:

Student: 2+9=11. Oh, I see, and in the third and eighth rows together there are 11 facts, also! And 4+7=11, and 5+6=11 too! So we have 5 pairs, and each pair adds up to 11. But I need to know multiplication to figure out the total: I have to multiply 5 by 11.

There are several easy ways to multiply a number by 11. Try to come up with a way or to of multiplying by 11. Does you way work for every number, or just for some numbers?

Mentor: Multiplication by 11 is very easy, especially for the first 10 whole numbers. Try to do the first few by adding 11 to each other or by counting cells, and see if you can find a pattern.

Student (using graph paper and marker and counting cells):

11*1=11

11*2=22

11*3=33

11*4=44

Oh, that's easy! It works all the way up to 9:

11*9=99

Mentor: There is a trick for multiplying larger numbers by 11 also. Anyway, do you see how counting by pairs and multiplying made computing 1+2+3+4+5+6+7+8+9+10=... easier?

Can you find a trick for multiplying larger numbers by 11?

Student: Yeah, every pair added to the same number. Then I just had to multiply that number, 11, by the number of pairs, 5, to get 55.

Mentor: Do you think you can use the same trick to find what 1+2+3+...+98+99+100 is?

Student: Why not? Let us see: the first plus the last is 1+100=101, then 2+99=101, then 3+98=101... There are 100 numbers, so there will be 50 pairs, each adding up to 101. So the total is 101*50. How do I figure that out?

Mentor: Think about it for a while. How many groups of fifty will be there in the total?

Student: 101, of course.

Mentor: Would it be easier if there were 100 groups of fifty?

Student: Yeah, then we would just have fifty hundreds: 5000. Oh, I see: we have fifty hundreds and fifty more, so it is 5050.

Mentor: The numbers 1, 2, 3, 4, ... is an example of an arithmetic progression. Here are more examples:

2, 4, 6, 8, 10, ...

5, 10, 15, 20, 25, ...

7, 10, 13, 16, 19, 22, ...

51, 61, 71, 81, 91, 101, ...

10, 5, 0, -5, -10, -15, ...

What do you think is common for all these sequences?

Student: I see some common things. The first one is just counting by 2's; the second one is counting by 5's; the third one... Wait... The third one is counting by 3's, starting from 7. In all of them you start from some number and keep adding another number.

Mentor: Would you like to construct your own arithmetic progression?

Student: Yes! I can start from any number and keep adding any number, right? Can I add halves?

Mentor: Of course!

Student: I am starting from 5 and adding halves:

5, 5 and a half, 6, 6 and a half, 7, 7 and a half,...

Mentor: If you want, you can write "and a half" as a point and 5 after a number. It reads "five-tenth" which is the same as a half:

For example, you can write "six and a half" as 6.5

Student: So my arithmetic progression looks like that:

5, 5.5, 6, 6.5, 7, 7.5, 8...

Mentor: The way you used to find the sum 1+2+...+100 will work for any arithmetic progression. Try, for example, to find the sum of the first ten numbers in my first progression (2, 4, ...), first by hand and then using the trick about pairs.

Student (adding using counting blocks): 2+4+6+8+10+12+14+16+18+20=110 It takes a long time to add all these numbers! Let us use the trick: the first plus the last here is 2+20=22, then 4+18=22, and so on. We have five pairs, because there are ten numbers, so we need to do 22*5. How can we compute that?

Mentor: There are many ways, of course. We can count 5's, for example.

Student: It is easy to count by 5's. We need twenty and two 5's. I know that ten 5's is fifty; then twenty 5's is a fifty and a fifty, which is a hundred. And we have two more 5's, which is ten, so we have one hundred and ten. The same thing I got by adding all the numbers!

Mentor: Do you think the trick will work for any arithmetic progression?

Student: Yes. It is not a trick, though, because it has a good explanation.

Create your own arithmetic progression. Find its sum using several different ways. Which way do you like better and why? Is one way always better for you, or would you prefer to use different ways with different progressions? Have you ever run into arithmetic progressions in life? Where? Look for them and see how many you can find. Ask other people to help you look.

Mentor: You can try to write this explanation down, so that you do not forget. Let us just express the trick, or rather the rule, in the language of algebra. Tell me again how you find the sum of an arithmetic progression?

Student: To find the sum of the arithmetic progression, you add the first and the last numbers in it, and then multiply by the number of pairs in it.

Mentor: Now, if you were to tell this rule to someone new to the idea, he may be wondering how to find the number of pairs.

Student: OK, to find the number of pairs, we take the number of numbers in the progression and find a half of it, of course.

Mentor: What if there are five or seven or any other odd number of numbers in the progression?

Student: Can you use the rule then?

Mentor: Let us try, with five numbers in the progression 1+2+3+4+5. We have two pairs 1+5 and 2+4 that both add up to 6, and then the middle number, 3.

Student: Well, can we count 3 as a half of a pair? Three is a half of six, after all.

Mentor: Of course! So, for five numbers, we have two and a half pairs, and each pair adds to 6.

Student (checking the arithmetic): It works! So our rule works even if the number of numbers in a progression is odd.

Mentor: Now let us translate your rule into the algebraic language. It will take less space this way. First, I would like to color it:

To find the sum of an arithmetic progression, you add the first and the last number of the progression, and then multiply by a half of the number of numbers in the progression.

Each word or words that I colored differently will translate into a part of the algebraic "phrase." I will translate it according to the traditions of the algebraic language. It is quite logical. Capital S is used for "sum" and some letter from an alphabet, say A, is used for numbers in the progression. A₁ can be used for the first number, A₂ for the second number, and so on, and A_L for the last number. N can be used for "the number of numbers." The rest of the colored words are translated into the algebraic "grammar":

S = (A₁ + A_L) * 1/2 * N

To find the sum of an arithmetic progression, add the first and the last number of the progression, and then multiply by a half of the number of numbers in the progression: S = (A1 + AL) * 1/2 * N