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 the number of terms in a sequence. So whenever you are asked to find the number of terms in a sequence the first thing we have to do is identify what kind of sequence we are looking at. So by looking at that we need to see some sort of pattern between the terms and figure out exactly what's happening.
So right here we are dealing with 7/3, 13/3, 19/3 so on and so forth and we are trying to figure out how many terms are on this thing. First thing we need to do is figure out is this an arithmetic sequence or not so we are going from 7/3 to 13/3. In order to do that we have to add 6 to a numerator, denominators are the same so we are really just adding to get from 7/3 or 13/3. See if we are doing the same thing to get to the next term if we are, it’s an arithmetic sequence.
Adding to 6/3 we end up at 19/3 so in fact we are having an arithmetic sequence, so therefore we know our general term of an arithmetic sequence will hold. So the general term a sub of n is equal to a1 plus n minus 1 times d.
So what we actually have here is we know our first term, we know our last term and we know our difference. We don’t know the number of terms. So we can plug in everything and then just solve for n. a sub of n is the nth term. We don’t know what term number it is but we do know what the nth term is and that’s the last term in this sequence just to be 85/3.
This is going to equal to a1, a1 is just our first term, 7/3, plus we don’t know what n is we don’t know how many terms are and times d. A common mistake that I’ve seen in this problem in finding d is knowing that you are going from 7 to 13 and 13 to 19 just calling d6 but it’s actually 6 over 3.
So our difference is really 2 not that 6 that’s just the numerator so this is just going to be times 2. Now we just have a linear equation which we have to solve. A number of different ways to solving this out we could multiply by 3 to get rid of the denominators we could distribute the 2, what I’m actually going to do is just subtract this 7/3 over to the other side, so 85 minus 7 is 78 over 3, this is just going to be equal to n minus 1 times 2. 78/3,I know that 78 is divisible by 3 because 7 plus 8 is 15 whenever your numbers add up to a multiple of 3 you have it's divisible by 3 but I don’t know what exactly it is, plug in the calculator turns out to be 26.
So this is then 26 is equal to n minus 1 times 2, dividing by 2 we get 13 is equal to n minus 1, add one to finish it up and we have found out that we have 14 terms in this sequence. Basically by using the fact that we have an arithmetic sequence here and plugging in information into our general term, we are able to solve it out to figure out how many terms we had in the sequence.
Unit
Sequences and Series