Unit
Sequences and Series
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!
To unlock all 5,300 videos, start your free trial.
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!
One other application of the arithmetic series is when we actually know what the sum is going to be and we want to figure out how many terms are needed to make that sum.
So as always we're going to go to our formula to figure out exactly how this works and our formula for our arithmetic series is s sub n is equal to n over 2 a1 plus an. And I know this is an arithmetic series because going from 5 to 7 we add 2 and then going from 7 to 9 I add 2 again, so there's that same difference, therefore it's an arithmetic series.
For this particular equation, we need to know the first term, last term and the number of terms. We're being asked how many terms are, so we obviously don't know what n is. We do know what a1 is, but we also don't know what a sub n is, so we have two unknowns which means we're going to have to mix things up.
We know the general term, so we can plug the general term in for a sub n to make sure that this equation doesn't depend a sub n, it instead depends on the difference.
So easy substitution, a sub is equal to n over 2, 2a1 plus n minus 1 times d. So same formula just a different way of writing it. Now let's plug in the information that we know. We want the sum to be 572, so that's our sum plug that in, 572, is equal to n over 2, n is the number of terms that's what we're looking for, so that's going to stay as n, 2 times a1, a1 is the first term which is 5, 2 times 5 is going to be 10 plus n minus 1 times d, our difference which is just going to be 2.
So we now have an equation which we need to solve for n. So let's take a look at this. I'm going to actually go ahead and distribute this n over 2 in because both terms it has to go to have a factor of 2, so we're going cancel things out to make our numbers a little smaller. So what we have then is 572 is equal to, this goes in leaving us with 5n and then plus here our 2s cancel, so we're left with n times n minus 1. Distributing this out 572 is equal to 5n plus n² minus n.
Combine like terms bring everything to one side we actually have a quadratic here what we wend up with is 0 is equals to n², 5n minus n this is just going to become plus 4n minus 572.
This one is a little bit harder to factor, 572 is a pretty big number, we could do the quadratic formula, but I'll just tell you right now that 572 is 22 times 26 which has a difference of 4 so what we end up with is n, n 122, oops, and the other is 26. We want to end up with a positive term, so our 26 is bigger meaning our 22 is negative, so that leaves us with n is equal to either -26 or 22.
Thinking about what we're solving for n is the number of terms in this series, so we have a negative number and a positive number. It doesn't make any sense that we have a negative number of terms not really, so we can dismiss that as our answer and leaving us with 22 terms in this series to add up to the sum that they wanted.
So just another application of using our arithmetic series equation in this case adding up a bunch of number in order to equal a set sum, how many terms are involved.