Learn math, science, English SAT & ACT from
highquaility study
videos by expert teachers
Thank you for watching the preview.
To unlock all 5,300 videos, start your free trial.
Geometric Sequences  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!
Using the general term for a geometric sequence can actually give us more than just what our general term or our specific term number is. For this example what we have is two specific terms inside of a sequence and we're trying to find the general term, so we don't know the first term per say and we don't know two consecutive terms, so we do know two terms and from that we have enough information to find out what our general term is or any term in the sequence for that matter.
So the trick for this is using our general term which is a sub n is equal to a sub 1 r to the n minus 1. And I have two pieces of information, we know that our second term is 6 and our 4th term is twothirds so from that what we can do is make a couple of equations and we can pair them up using elimination or substitution or something in order to solve that out.
So the first equation we can go to is a sub 2 is equal to 6 and again remember we have to distinguish between what is the term number and what the actual term is. So this is saying a sub 2 is 6 which tells me my second term is 6, so n is 2 and a sub 2 is 6, so from there we get 6 is equal to a sub 1, 6 is equal to a sub 1 times r to the 2 minus 1 or r to the first.
The other equation we get is 2/3 is equal to a sub 1 r to the 4 minus 1, r to the third. So we have two equations we now need to put them together somehow. A couple of ways of doing this, we could solve for a sub 1 and set them equal to each other because the first term in the sequence has to be the same, the first term of any sequence is going to be whatever it is so this terms has to equal that term.
We also could divide one by the other, which is typically a different skill that some people haven't seen before but basically just like any system, you can add, subtract them, you can also divide them and in this case it works out because we really only have one term per side it makes our life fairly simple, but if you haven't seen that I'm going to stay away from it and just do it by substitution.
So from here I know that a sub 1 is equal to 6 over r. From over here I know that a sub 1 is equal to 2 over 3r³. Using substitution we know that these two a sub 1s have to be equal to each other so let's go over here and solve this out. So we know that 6 over r is equal to 2 over 3r to the third, cross multiply 18r³ is equal to 2r, put everything to one side, 18r³ minus 2r is equal to the 0, factor out and solve. So let's factor out 2r and we're left with, it becomes 9r² minus 1 equals 0 leaving us with r is equal to 0 or r is equal to plus or minus 1/3 if you want to factor this out one more time, just the difference of squares we know how to factor that.
So we are left r is 0 or r is equal to plus or minus 1/3. R is 0 doesn't make any sense, if our r is 0, our terms will all be the same so we somehow created a solution or r is equal to plus or minus 1/3. This case a little interesting because we have two terms that are two spots away, so if our rate was positive or negative it would matter because if they're both positive obviously it says positive, but if it's negative, we multiply by one rate it becomes the opposite sign and again the other becomes back to what it is.
So for this particular example r could be either plus or minus 1/3, we don't have enough information to really distinguish it. We know that numerically it has to be 1/3, but that negative could be in there or it might not be. So we know that our rate is either plus or minus 1/3. Let's just say it's 1/3 except that we have this potential for an extra solution. Now all we have to do is find out what our first term is.
Well we have two equations. Just like any systems of equation, all we have to do is go back n and plug information in to find the other term, so we could go to either equation, I'm going to go to this one just because it's a little bit simpler, so we get 6 is equal to a1 times 1/3 multiplied by 3 and we find out that a1 is equal to 18.
So we have our a1 and we have our rate, to find the general term all we have to do is put those things back together and we get an is equal to a1 which we know is 18, times r to the n minus 1 and again in this case the r could be plus or minus we don't really have enough information to figure that out.
So taking two terms from a geometric sequence, find the general term, make a system of equations, solve it out and plug it back in to find your other variable, throw it all together to make your general term.
Please enter your name.
Are you sure you want to delete this comment?
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.
Concept (1)
Sample Problems (12)
Need help with a problem?
Watch expert teachers solve similar problems.

Geometric Sequences
Problem 1 9,129 viewsGiven 27, 9, 3, 1 ...
find a_{n} and a_{8} 
Geometric Sequences
Problem 2 7,735 viewsGiven a geometric sequence with a_{2} = 6 and a_{4} = ⅔.
Find a_{n} 
Geometric Sequences
Problem 3 1,204 views 
Geometric Sequences
Problem 4 1,121 views 
Geometric Sequences
Problem 5 1,124 views 
Geometric Sequences
Problem 6 1,171 views 
Geometric Sequences
Problem 7 1,056 views 
Geometric Sequences
Problem 8 1,078 views 
Geometric Sequences
Problem 9 1,166 views 
Geometric Sequences
Problem 10 1,074 views 
Geometric Sequences
Problem 11 1,062 views 
Geometric Sequences
Problem 12 1,053 views
Comments (0)
Please Sign in or Sign up to add your comment.
·
Delete