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Arithmetic Series  Problem 2
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Finding the partial sums of an arithmetic series. So for this example what we are trying to do is add this arithmetic series up. I know it's arithmetic because in order to get from 3 to 3/2, I have to add one and a half and to get from 3/2 to 0 I have to add one and a half as well, so I'm adding one and a half each time until I get up to 30.
So we know what our partial sum equation is to find our sum is. S of n is equal to n over 2 a1 plus an and in this particular example we know what our first term is, that's going to be 3 and we know what our last term is and that is 30. The problem is that we don't know what n is, we don't know how many terms are in this series, so what we have to do is we actually have to go back and figure out how many terms there are. The way we do that is just by using our general term.
So what we can do is we can go to our general term a sub n is equal to a1 plus n minus 1 times d and we have to figure out what n is. A sub n is our last term it's 30, a sub 1 is our first minus 3 plus n minus 1 times our difference, our difference is just what we do to get from one to the other which is 3/2.
So add 3 to the other side, 33 is equal to n minus 1 times 3/2 multiply both sides by 2/3, 33 over
3 is 11 times 2 is 22, so we end up with 22 is equal to n minus 1 which tells us that n is 23. So we have 23 terms in this sequence, plugging that in 23 over 2 is equal to s sub 23. Plug this into our calculator, 30 plus 3 is 27 times 23 divided by 2 is equal to 310 and a half.
So finding a sum of a series where we don't know how many terms are, all we have to do is plug in our formula and then go back to our general term in order to figure out how many terms are and then plug that in to finish up our equation.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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