دنباله حساب

سرفصل: بخش ریاضی / سرفصل: مسائل کلمه / درس 17

دنباله حساب

توضیح مختصر

  • زمان مطالعه 8 دقیقه
  • سطح متوسط

دانلود اپلیکیشن «زوم»

این درس را می‌توانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید

دانلود اپلیکیشن «زوم»

فایل ویدیویی

متن انگلیسی درس

Arithmetic Sequences

Arithmetic sequences. One special kind of sequence is called an arithmetic sequence. An arithmetic sequence is one in which we add the same constant to get from each term to the next. So if you look at this sequence, you notice that to get from each term to the next, we’re simply adding seven.

That’s how we go from one term to the next. So, we can continue this pattern forever, just continually adding seven to get the next number. Notice first of all, a really big idea. Any evenly spaced list is an arithmetic sequence. So, this is powerful.

One special case is consecutive multiples of an integer P we would add P to get from each term to the next. Consecutive odd numbers consecutive even numbers or consecutive integers are also special cases of arithmetic sequences. Here’s an arithmatic sequence. It would be helpful to find a formula for the nth term, so that, for example, we could find the value of a much higher term in the sequence.

The test likes to ask about this. Remember, for the algebraically defined sequences that we saw in the previous video, it’s very easy if the test said find the 80th term in this sequence. All we’d have to do is plug in. Well, we’d like a formula that would allow us to do that for arithmetic sequences. I’m gonna analyze the individual terms in the following way to look for a pattern.

So I’ll focus on this particular sequence, the sequence that starts with five. And then we add seven each time. So the verse term is five the second term is 12 I’m going to write that as five plus seven we’ve added seven once. To get the third term we add seven again so I’m going to write that as five plus seven plus seven.

Then the fourth term we add seven again, then add seven again and so what we have here, this stream of sevens. We have five plus a number of sevens. Notice that the number of sevens in each row, the number of sevens that we’re adding, is one less than the index number. It’s one less than the number on the list.

You need to turn the number seven’s number added one less than the index of the term. This is a big idea. Thus, the nth term AsubN would be the starting term plus n-1 individual 7’s. Of course, a compact way to add n minus one individual sevens would simple be to multiply 7 (n-1). So that means that the nth term of this sequence would be 5+7 (n-1).

Now, we can generalize a bit. First of all, the starting term in general is a sub one. We’re familiar with that from the last video. The fixed amount we add each time to get each term is called the common difference. Because if we subtract any two adjacent terms we get this same number. We symbolize the common difference with the letter d.

The nth term of an arithmetic sequence with an initial term of a sub one, and a common difference d is a sub n equals a sub one plus n minus one times d. That is a very powerful formula. Please do not simply memorize it. Please understand the argument that led to this formula, so that you understand it more deeply.

But we’ll be using this formula quite a bit. Notice that one context in which evenly spaced integers arise is the set of all positive integers that, when divided by one number, give a fixed remainder. For example, this sequence. The sequence that we’ve been looking at. Is the set of all integers that, when divided by seven, have a remainder of five.

That’s what all those numbers have in common, when we divide by seven, we have a remainder of five. Notice that the remainder is the first term, a sub one. And the divisor is the common difference, that will be valuable so hold on to that thought. Here’s a practice problem, very simple practice problem, we have an arithmetic sequence, find the 41st term of this sequence.

Pause the video, and work on this. Okay well the first term is 14 and obviously we’re adding nine each time. That’s how we’re advancing so the common difference is nine. So we can just plug into the formula this is the general formula if we want the 41st term we plug in N equals 41 so we get 40,14+9 (40). 9 40 of course is 360.

Add 14, we get 374. 374 is the 41st term on the sequence, the 41st term on the list. Here’s another practice problem. Pause the video and then we’ll talk about this. Let S be the set of all positive integers that, when divided by eight, have a remainder of five.

What is the 76th number in this set? Well, remember that we said that the remainder five, that’s gonna be the first term on the list. Because, of course, the lowest number on the, in this whole set is gonna be five itself. If we divide five by eight, we get a quotient of zero and a remainder of five.

And then eight will be the common difference, because we’ll be adding eight each time to get the new numbers. So we have our starting term. We have our common difference. That’s our general formula. We plug in n equals 76 for the 76th number.

We get 5+8 (75). For that eight times 75 we’ll just do doubling and halving. So it’s four times 150. Four times 150 is clearly 600, plus five is 605. And that is the 76th number on the list. Here’s another practice problem.

This one’s a little bit different. Pause the video, and then we’ll talk about this. Okay. Here it turns out that the first term is not known. The common difference is not known. We’re just given on two random terms somewhere on the list.

And we wanna find another term. Well think about it this way. Let the initial term equal b. And let the common difference equal d. Well then certainly its true that A sub three would be B plus 2D and that’s 17. The 19th term would be B plus 18D and that equals 65 and what we have here are two equations with two unknowns and in fact its very easy to solve.

We’re just gonna subtract the first equation from the second equation. Subtract. We get 16d equals 48. Divide by 16 we get d equals three. And then plugging into a3 we get, that the initial term is 11. So now we have the initial term.

The common difference. We can set up the general formula. The 10th term will be initial term 11 plus common difference three times ten minus one, which is nine. Three times nine is 27. We add that, we get 38.

And 38 is the 10th number on the list. An arithmetic sequence is one in which the terms have a common difference. Any evenly spaced list is an arithmetic sequence. Other examples include consecutive multiples of a number, consecutive odds or evens, and numbers which, when divided by the same divisor have the same remainder. The nth term of any arithmetic sequence is given by the formula a sub n equals a sub one plus d times n minus one.

مشارکت کنندگان در این صفحه

تا کنون فردی در بازسازی این صفحه مشارکت نداشته است.

🖊 شما نیز می‌توانید برای مشارکت در ترجمه‌ی این صفحه یا اصلاح متن انگلیسی، به این لینک مراجعه بفرمایید.