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!
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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!
Finding a specific term in an arithmetic sequence when we don’t know the first term or the difference. So what we have here is two terms the 11th term and the n19th n term and we are asked to find the 1001 term.
There are two ways if doing this problem. There’s one that I’m going to do and there’s one that I’m not going to do and I’ll just give the brief overview of the one that I’m not going to do and the reason I’m not doing it is kids kind of get confused when they try this one. And what we can do is actually sort of reset everything and call your lower term number your first term.
So make this a sub of the bizarre one we make this a sub of bizarre 8 and then you could then find your difference between those and use this as your point of reference.
In general that creates a lot of confusion you tend to get off by one when you figure out what your new bizarre term numbers are. So I tend to skip that, I’m not going to do that way if you figured out it works for you go ahead and use it but I’m going to do the safe way which I know students don’t tend to mess up as much.
And the way we do that is basically by making a system of equations. So what we can do is using our general term a's n is equal to a1 plus (n minus 1) d what we can do is make an equation for both of these pieces of information.
So a's of 11 is 52 the 11th term is 52 that equation is going to be 52 is equal to a1 which we don’t know, plus this is the 11th term so 11 is going to go in here. To save some space I’m going to skip this step and basically say 11 minus 1 is 10. If you want to write to 11 minus 1 in feel free but I’m going to skip the step and just write this as 10d.
Be careful when you are putting these in not to confuse a sub of n with n, a sub of n is what the actual term is n is what the term number is. The term number is the little subscript, what the term is is the big number. I often see students switching these or writing the same number twice or whatever it is. Make sure you distinguish between what the term is and the term number.
We now need to do the same exact thing for the other piece of data. The 19th term is 92 again making sure you distinguish between the term number and what the term is, the term is 92 so that’s going to go by itself over here, is equal to a1 plus plugging in 19 for our n, the 19th term again I’m going to skip the step of writing out 19 minus, 19 minus 1 we know to be 18 and then times d.
So what you notice now is we have a system of linear equations. Two systems, two unknowns, the cool thing about this system is that a1's are always going to be by themselves so we can always just do a simple elimination and our a1 are going to disappear. All we have to do is subtract.
52 minus 92 is -40, a1 minus a1 those cancel, 10 minus 18 is -8d. Solving for d, divide by -8 we end up with d is equal to 5. So we found our difference, just like any other linear system to find the other unknown all we have to do is take this and plug it back into either of these equations. I like 5 times 10 a lot more than 5 times 18 so I’m going to plug in to this top equation but you could just as easily plug it into the bottom one you should get the same answer. So we take this plug it in here and we end up with 52 is equal to a1 plus 10 times 5.
10 times 5 is 50 subtract it over and we find that a1 is equal to 2. So we have our 5 for our d, we have a 1, so let’s go and make our general term for this equation as a whole, which is basically going to be a sub of n is equal to a1 just 2, plus n minus 1 times d which is 5. So we found the general term what the problem is asking for is the 1001. All we have to do now is plug in 1001, 2 plus 1001 minus 1 is just going to be 1000 times 5, 1000 times 5 is 5000 plus 2, 5002.
So we are able to solve this system given that we knew two terms we made a system if equations solve for d and for a1 went back and plugged it in.
I will have to mention that I did make one assumption which I forgot to sort of specify at the beginning. This only works because we know it has an arithmetic sequence. If this wasn’t arithmetic if we weren’t adding something each time these equations will necessary hold true.
So in my writing up I should have said knowing an arithmetic sequence with these, find this up I forgot I apologize, but hopefully you’ll see the process and you’ll see somewhere in your written word problem what exactly is going on.