Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. Arithmetic sequences in series The definition of an arithmetic sequence an arithmetic sequence is a sequence in which each term after the first differs from the preceding term by a constant amount. The difference between consecutive terms is called the common difference of the sequence, the general term of of an arithmetic sequence, the NTH term, or the general term of an arithmetic sequence with the first term A1. And the common difference D is a sub N equaling a sub 1 + n -, 1 * d So if we looked at an arithmetic sequence, an example would be something like 13579, etcetera, where this is the first term, the second term, the third term, the fourth term, the fifth term. And so if we thought about what did we do to get from the first term to the second, we added a two. To get from the 2nd to the 3rd, we added a two. To get from the third term to the fourth term, we added a two. That common difference is 2. And if we thought about this, we could think of the first term as one and the next term is 1 + 2 and the next term is 1 + 2 twos. And the next term is 1 + 3 twos and the next one is 1 + 4 twos. And if we look at this and we think about, the first term didn't have any twos added, the second term had 1-2 added, the third term had two twos added, the fourth term had three twos added, the fifth term had four twos added. So if we thought about coming out here far enough, the ace of NTH term would be whatever the first term was, plus one less than the quantity of N So if it had been five, we needed four. If it's N, we need n -, 1. If we look at doing a sum, our S sub N would be a sub 1 + a sub 1 + d for our second term plus a sub one plus 2D for our third term plus a sub one plus 3D plus... And way down here at the end, we're going to have our a sub NTH term -2 D. And then we're going to have our a sub NTH term minus D. And then we're going to have our a sub NTH term. So we add D all the way up. And then when we get to the end, we don't really know what the end is. But to get from the end of the term, right before it, we would have subtracted D and then we would have subtracted another D. Now I'm going to write right below it S sub N again. But this time, instead of going in ascending order, I'm going to go in descending order. So I'm going to start with my a sub N and then I'm going to have my a sub n -, d term and then the a sub n -, 2 D term plus... dot plus a sub one plus 2D at the end and then a sub 1 + d + a sub one. If we thought about then taking those two equations and adding adding them together, we'd get 2 S sub NS on the left. A 1 + a N would give us a 1 + a N If we look at this next one, A 1 + d plus A n -, d is going to give us a 1 + a N. If we look at the next one, A1 plus 2D, A n - 2 D is going to give us a one plus an. And if we do the same pattern, we see that every single one of those terms is going to give us a sub 1 + a sub N. Now each of those terms have an A sub 1 + a sub N in common. So if I factored that out, I'd get a 1 + 1 + 1 plus. Well, how many ones are there? There are north of them, so we'd get 2S sub N equaling a 1 + a N times N Then we're just going to divide by two. So we get SN and we can multiply in any order we want. So I'm going to put that quantity towards the end so that now we have an equation for any time we want the sum of an arithmetic sequence. The sum of an arithmetic sequence is just n / 2 times a 1 + a N or N times a 1 + a N all over 2. So the sum of the first N terms of an arithmetic sequence, the sum S sub N of the first N terms of an arithmetic sequence is given by S sub N equaling n / 2. The quantity a 1 + a N in which A1 is the first term and AN is the NTH term. Let's look at some examples. If I have a sub one equal -7 and D equal 4, let's find our first three terms. If we have a sub two, we'd have -7 plus our term minus our one times our difference. So we'd have -7 + 1 * 4 or -3 our a sub third term would be -7 + 3 - 1 * 4 or -7 + 2 * 4, which is one are a sub four negative 7 + 4 - 1 * 4 negative 7 + 3 * 4 or five. Now, if we thought about writing these out in a list, we would hopefully be able to see that here we just subtracted 4 and then we subtracted 4 again. And then we subtracted, oh, we added four, we added four, and we added four and we added four. So the next one we would add 4 to get 9, etcetera. This next one, a sub one, is -9 a sub two. It's a recursive equation, which means it's based immediately on the term right before it. So our A sub one term was -9 + 6 or -3. Our a sub three term is going to be based off of the a sub two term. So our A sub two term +6 is going to give us 3. Our A sub four term is going to be based off of our A sub three term and we're going to add 6. If we make this into a list -9 negative 339, we can see we added six and we added six and we added six. And if we added six again, we could get the next term, the next one. We want to find a of 50 given the fact that A of one is 7 and our just difference, common difference is five. So if we want A of 50, we know that we're going to have our A sub first term plus our 50 -, 1 times our D. Well, our A sub one term was just seven. So we're going to get 7 plus 49 * 5 five carry four 245 + 7, so 252. If we wanted our A of 60 term, our A of 60, our first term being 35 + 60 -, 1 times -3 So 35 + 59 * -3 negative 7 two 177 + 35 S -142 seven carry 2. Yep, 7/3 negative one -5 on this kind of problem, what it's asking us to do is it wants us to come up with the NTH term. So we can see here that we subtracted 4 and we subtracted 4 and we subtracted 4. So our a sub N term is going to be the first term plus our n -, 1 term times -4. Now we can actually simplify this up. So we'd get 7 - 4 N +4 or 11 - 4 N Then we're actually asked to find the 20th term. So if we have a sub twenty, we put in 11 - 4 * 20 or 11 - 80. So -69 if we have a sub one equal 4 and a sub N equaling A n -, 1 + 3, we can see that we're going to have four and then seven and 10 and 13. So our A sub N term here is going to be our first term plus our n -, 1 term times our common difference, 4 + 3 N -3 or 1 + 3 N If we wanted our 20th term, 1 + 3 * 20 or 61. We want to find the sum of the 1st 20 terms, so we need to know our A sub one, which we do is 4, but we also need to know our A sub 20 and our A sub 20. We're going to get by 4 + 20 -, 1 times our common difference, which here is 6. So we're going to get 4 + 19 * 6 and that's going to give us 19 * 6 + 4 a 118 four carry 5. Yep. And then we're just going to use the formula that says our S sub N equal our N in this case 20 / 2, our first term plus our last term. So 10 * 118 + 410 * 118 + 4 is going to give us 1220 for our 1st 20 terms. That shouldn't have been an end back there. It should be a 20. If we look at a of 21, we want the sum of the integers between 21 and 45. So we know that a sub one is 21 and we know our last term is 45, but we've got to actually figure out what the N is going to be in this one. So we need to figure out if our a sub N is 45. We know that 45 would equal our a sub one term plus our n -, 1. And if they're even integers, our common difference is going to be two. So to solve this, if we take 45 and subtract 21, we're going to get 24 equaling n -, 1 * 2. If I divide by two, I get 12, so N is going to be thirteen. So if we thought about listing this out, if we have 21 and then 20. So the next example wants the even integers between 21 and 45 S 22242628... To 44 we want the even integers. So my a sub one is going to be 22. My a sub N is 44. But we need to figure out how many terms there really is. We need to figure out what that N is. So our a sub N equals our a sub 1 + n -, 1 times the common difference. If they're even integers, our common difference is 2. So 44 - 22 would be 22 equaling n - 1 * 2. If we divide by two, we get 11 equaling n - 1 or N is 12. So now to find the sum, we just use our formula that says n / 2 of the first term plus the last term. So we'd have 6 * 66 or 6 carry three 396. The next type of problem, they want us to write out the first three terms and the last term and then figuring out how to find the sum. So the first term, if I stick in one I get -2 + 6 or 4. The next term when I stick in two I get -2 + 6. If I stick in two I get -4 + 6. A sub one -2 * 1 + 6 negative 2 + 6 would be 4, a sub two -2 * 2 + 6 negative 4 + 6, which would give us 2A sub three -2 * 3 + 6 negative 6 plus 60A sub 40. The last term -2 * 40 + 6, negative 80 + 6 or -74 So we want to add up all these terms. When we look at this, we can see that our common difference here is going to be -2 we're going down to each time. So our S of 40 is going to equal the 40 terms over to our first term plus our last term. So 40 / 220 times the 40 + -74 is -70 so negative 1400, our last type of problem. They're going to give us a picture, and they want us to find different summations of A to the 14th plus B to the 12th. So our first graph is an if we have a sub one. Looking at this graph as 1A, sub two is -3. A sub three is -7 Literally this is my N. So a sub one was the number one. A sub two was the number -3. A sub three was the number -7 So we can see that our common difference here is going to be -4 So if I want the 14th term, it's just the first term plus 14 -, 1 times our -4. So 1 + 13 * -4 one minus 52 or -51 over here, my B sub one. This is B sub N and there's my end. So my B sub one is 3, my B sub two is 8, my B sub three is 13. If we thought about connecting these points, we could see that this is going to be a line. Our common difference for this one, we're adding five each time. So if I want my 12th term, it's the first term plus 12 -, 1 times the common difference. So 3 + 11 * 5, three plus 55 or 58. So if I have -51 + 58, the sum there would be 7. Thank you and have a wonderful day.