 ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# Arithmetic Series - Problem 1

Carl Horowitz ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Finding the partial sum of an arithmetic sequence; so what we're looking at right here is we have an arithmetic sequence we know it's arithmetic because each time we are adding 4 to get to the next term and we are trying to find s of 40 which is the sum of the first 40 terms.

So we have a formula that is s of n is equal to n over 2 a sub 1 plus a sub n. We know that the general term a sub n is equal to a1 plus n minus 1 times d, so we can do is actually substitute that in here. And why I'm doing that is this particular equation we need to know the first term and the last term. For this particular example, we don't know the last term, so we have to figure that out. So by plugging in what our general term for a sub n, we can rewrite this equation a sub n is equal to 8 over 2, 2a1 plus n minus 1 times d.

So what I've done just there is I've changed it from having to know my last term to just having to know my difference which by looking at this I can know. So all we have to do now is plug in some information.

We are looking for s of 40, the sum of the first 40 terms, that means n is 40, we can just plug that in, 2 times a1, a1 is just our first term which is just going to be 3, plus n minus 1, n is the number of terms, so this is just going to be 39 and times our difference, our difference is what we do from one term to get to the next, so our difference here is just 4.

So what we have is this fairly elaborate equation, but it's all numbers, we can just plug into our calculator starting at our parenthesis 39 times 4 is 156 plus 6 is 162 times 40 over 2, so times 20 which leaves us with 3240.

So by adding up our first 40 terms, we end up with our sum, using either one of our two summation equations either one would work, but for this one I find the second one to work a little bit better because we have the difference and not the last term.