Then we add 6 349 times, so we have 350 terms in this sum, so in this case, n is The first term just yet, so we're going to have, so we have 300, we have the first term and You have the 349 forĮvery time you added 6, so this is the first time you added 6, second time you added 6, all the way to the 349th time you added 6, so let me make it clear, this, this is, oh, actually, this is theģ49th time I added 6 to get to this, but we haven't counted Tempted to say 349 terms, but really, you have 349 plus 1 terms. So to go from negative 50 to 2,044, I have to add 6 to 349 times, so I add it once, I add it twice, and then this right over here, this is the 349th time that I'm adding 6, so how many terms do I have? Now, you might be So 6 goes into 20 three times, 3 times 6 is 18, subtract, 20 minus 18 is 2, bring down the 9, 6 goes into 29 four times, 4 times 6 is 24, subtract, 29 minus 24 is 5, bring down the 4, we have 54, 6 goes into 54 nine times, 9 times 6 is 54 and we are done. To get back to zero, then go up another 2,044 so I have to go up by 2,094, so if I'm going, if I'm adding 6 on every term, how many times do I have to add 6 to increase by 2,094? Well, let's just takeĢ,094 and divide it by 6 to figure that out. To get back to zero and then go up another 2,044. How far do I have to go from negative 50 to 2,044? I have to go up 50 just Reason I calculated this is I want to figure out Thing as 2,044 plus 50 or 2,094, and the whole That is a1, and this is our last term, 2,044, so that is our a-sub-n, so the other question is, well, what is n? How many terms do we actually deal with? And to think about that, we just say, well how many times do we have to add 6 to go from negative 50 to 2,044? Well, 2,044 minus negative 50, minus negative 50, well that's the same Our first and last terms are, we know this right over here. The average of the first and the last terms, times the number of terms that we're dealing with. Of variety on the screen, if we're taking the sum of, of the first n terms ofĪn arithmetic sequence or, if we're taking or if we're evaluating theįirst n terms of an arithmetic series, I could say, it's going to be the first term plus the last term divided by 2. We know that if we have, if we are taking the sum of, let me do this in a new color, just to have a little bit So we know how to take the sum of an arithmetic sequence. Each term is 6 more, is a constant amount more than the term before that. So we are dealing right over here, this sum is an arithmetic series. And so each successive term is just 6 more than the term before it. To get to that last term, we add 6 once again. Negative 44 plus 6 is negative 38, and we go all the way to here, we keep adding 6, and to go from 2,038 to 2,044, So the first term here is negative 50, and then we go to negative 44, so the second term is negative 50 plus 6, and then the third So let's work through this together, and let's just thinkĪbout what's going on. Negative 38, all the way, we keep adding all the way Have the sum negative 50 plus negative 44, plus
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |