Arithmetic progression
Código VBLB-E0012-I
VIEW:575 DATA:2020-03-20
Arithmetic progression is when we see a sequence of numbers that are constructed by adding, or subtracting a number.
For example.
1, 2, 3, 4, 5.
It is an arithmetic progression because we add one more always and we make such a progression.
1, 1 + 1, 1 + 1 + 1, 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1.
See that each time I move to a number I add up 1.
What if it was number two? We would have.
1, 3, 5, 7, 9.
That is now the sum of the number 2.
1, 1 + 2, 1 + 2 + 2, 1 + 2 + 2 + 2, 1 + 2 + 2 + 2 + 2.
See that we have the odd numbers, but if now instead of starting with 1, we start with 0, and add up to 2, we would have the following numbers.
0, 2, 4, 6, 8.
What would be even numbers now. See that only the starting number has changed.
0, 0 + 2, 0 + 2 + 2, 0 + 2 + 2 + 2, 0 + 2 + 2 + 2 + 2.
But what would be the use of this type of calculation?
Let's say that every day you post 2 images on your social media, how many days would it take to post 30 images?
Now, I could divide 30 by two and I would have 15 days.
But let's see how it does this using Arithmetic Progression.
Before there was no image, so my initial quantity is 0. Let's call it a0 (zero).
a0 = 0;
So I upload two images on the first day. Let's call the number of images on the first day a1 (the one).
a1 = a0 + 2;
Then on the second day I upload two more images.
a2 = a0 + 2 + 2;
So on the third day I put two more pictures.
a3 = a0 + 2 + 2 + 2;
Try to observe that we have 3 sums of number 2, could we not do a multiplication? Watch.
a3 = a0 + 2 + 2 + 2 = a0 + 3 * 2;
Note that the sum of 3 numbers 2 is the same as three times the number 2.
a3 = a0 + 3 * 2;
note that the number 3 appears in the index. See if we do with 4.
a4 = a0 + 4 * 2;
Now that we're going to do the magic, instead of calling it 3 or 4 or 5 we're going to put n, representing any number.
an = a0 + n * 2;
Observe carefully because this is where all the arithmetic progression is based. This n is the generalization for using any number, so we can calculate very large numbers.
Remember the example of the images. We will see in 15 days, in 15 days the "n" would be 15, and thus replacing we have.
a15 = a0 + 15 * 2;
Remember that there was no image, so a0 (at zero) is 0.
a15 = 0 + 30 = 30;
Then on the 15th, we would have 30 images posted.
But we are going on the 20th now. So we have.
a20 = a0 + 20 * 2 = 0 + 40 = 40
That is, on the 20th we would have 40 images posted.
Remember the generalization we made.
an = a0 + n * 2;
Let's say now that I put instead of two and put 3 images, the formula would be like this.
an = a0 + n * 3;
But now let's say that I put 5 images a day, so we would have this formula.
an = a0 + n * 5;
Now let's call the number of images "r", so the formula would look like this.
an = a0 + n * r;
So, now we can generalize all of our arithmetic progression.
How many images will I have in 10 days, I had no image and I put 10 images a day? The formula would be this.
a10 = 0 + 10 * 10 = 100;
so in 10 days i will have 100 images. But if from now on I put 5 images in another 10 days? Remember that now I already have 100 images, so my a0 (at zero) is 100. And now it's 10 more days.
a10 = 100 + 10 * 5 = 150;
So, now I will have posted 150 images. See that before in the first 10 days I posted 100, and in the other 10 days it added 50, and I would have 150 images.
This is very useful for large projects, and for computer programming. It is also used for game programming, in the amount of coins, or objects that the character obtains, and many other uses. Remember to do exercises to fix.
For example.
1, 2, 3, 4, 5.
It is an arithmetic progression because we add one more always and we make such a progression.
1, 1 + 1, 1 + 1 + 1, 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1.
See that each time I move to a number I add up 1.
What if it was number two? We would have.
1, 3, 5, 7, 9.
That is now the sum of the number 2.
1, 1 + 2, 1 + 2 + 2, 1 + 2 + 2 + 2, 1 + 2 + 2 + 2 + 2.
See that we have the odd numbers, but if now instead of starting with 1, we start with 0, and add up to 2, we would have the following numbers.
0, 2, 4, 6, 8.
What would be even numbers now. See that only the starting number has changed.
0, 0 + 2, 0 + 2 + 2, 0 + 2 + 2 + 2, 0 + 2 + 2 + 2 + 2.
But what would be the use of this type of calculation?
Let's say that every day you post 2 images on your social media, how many days would it take to post 30 images?
Now, I could divide 30 by two and I would have 15 days.
But let's see how it does this using Arithmetic Progression.
Before there was no image, so my initial quantity is 0. Let's call it a0 (zero).
a0 = 0;
So I upload two images on the first day. Let's call the number of images on the first day a1 (the one).
a1 = a0 + 2;
Then on the second day I upload two more images.
a2 = a0 + 2 + 2;
So on the third day I put two more pictures.
a3 = a0 + 2 + 2 + 2;
Try to observe that we have 3 sums of number 2, could we not do a multiplication? Watch.
a3 = a0 + 2 + 2 + 2 = a0 + 3 * 2;
Note that the sum of 3 numbers 2 is the same as three times the number 2.
a3 = a0 + 3 * 2;
note that the number 3 appears in the index. See if we do with 4.
a4 = a0 + 4 * 2;
Now that we're going to do the magic, instead of calling it 3 or 4 or 5 we're going to put n, representing any number.
an = a0 + n * 2;
Observe carefully because this is where all the arithmetic progression is based. This n is the generalization for using any number, so we can calculate very large numbers.
Remember the example of the images. We will see in 15 days, in 15 days the "n" would be 15, and thus replacing we have.
a15 = a0 + 15 * 2;
Remember that there was no image, so a0 (at zero) is 0.
a15 = 0 + 30 = 30;
Then on the 15th, we would have 30 images posted.
But we are going on the 20th now. So we have.
a20 = a0 + 20 * 2 = 0 + 40 = 40
That is, on the 20th we would have 40 images posted.
Remember the generalization we made.
an = a0 + n * 2;
Let's say now that I put instead of two and put 3 images, the formula would be like this.
an = a0 + n * 3;
But now let's say that I put 5 images a day, so we would have this formula.
an = a0 + n * 5;
Now let's call the number of images "r", so the formula would look like this.
an = a0 + n * r;
So, now we can generalize all of our arithmetic progression.
How many images will I have in 10 days, I had no image and I put 10 images a day? The formula would be this.
a10 = 0 + 10 * 10 = 100;
so in 10 days i will have 100 images. But if from now on I put 5 images in another 10 days? Remember that now I already have 100 images, so my a0 (at zero) is 100. And now it's 10 more days.
a10 = 100 + 10 * 5 = 150;
So, now I will have posted 150 images. See that before in the first 10 days I posted 100, and in the other 10 days it added 50, and I would have 150 images.
This is very useful for large projects, and for computer programming. It is also used for game programming, in the amount of coins, or objects that the character obtains, and many other uses. Remember to do exercises to fix.
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Tags
Arithmetic progression, mathematics, PA, progression, algebra