Finding the Sum of Arithmetic SeriesDate: 06/12/2006 at 21:03:24 From: Marissa Subject: Sum of an Arithmetic Progression Dr. Math, I was wondering what was the best procedure or method to find the sum of an arithmetic progression. The method I have been using seems so long and a little confusing. Can you explain it using this example? 4 + 10 + 16 + 22 + ... + 58 I seem to get lost in the steps at some point when you have to determine the amount of terms in the sequence. The steps I'm using are: 1. Find the sum of the first and last digits. 2. 2s=62+62+62+62+...+62 3. The difference between succeeding terms in this progression is 6. 4. I get lost somewhere along here. Date: 06/12/2006 at 21:57:03 From: Doctor Greenie Subject: Re: Sum of an Arithmetic Progression Hi, Marissa - From the work you show, I only know a little bit about the method you have been using. So let me go through the entire solution process that I use. (1) Strategy... The average of ANY group of numbers is the sum of them all, divided by how many there are: average = sum / how many We can turn this definition of average around to say that the sum of any group of numbers is their average, multiplied by how many there are: sum = average * how many We can use this formula to find the sum of any arithmetic sequence. We only need to determine the average of all the numbers in the sequence and how many numbers there are. (2) Average of the numbers in an arithmetic sequence... Because the numbers in any arithmetic sequence are equally spaced, the average of all of them is the average of the first and last numbers: average = (first + last) / 2 (3) Number of numbers in an arithmetic sequence... (a) If we subtract the first number in the sequence from the last, we find how far it is from the first number to the last number. (b) If we divide the difference found in step (a) by the common difference between terms, we find out how many terms there are in the sequence after the first one. Note this means that if (for example) the result of the division here is 11, then the number of terms in the sequence is 12. Let's look at your particular example, following the steps outlined above. 4 + 10 + 16 + ... + 58 (2) Average of all the numbers in the sequence... average = (4+58)/2 = 62/2 = 31 (average of first and last numbers) (3) Number of numbers in the sequence... 58 - 4 = 54 (The last number is 54 greater than the first.) 54/6 = 9 (Since the common difference is 6, the last number, being 54 more than the first, is 9 terms after the first term; the number of terms is 9 + 1 = 10.) (1) Sum of the numbers in the sequence... The sum is average * how many = 31 * 10 = 310 I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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