 In this video we'll talk about geometric sequence. A geometric sequence, remember sequence is a pattern of terms and when we had arithmetic we were adding the same thing over and over again, we're going to be multiplying by a common value to get to each successive term in geometric sequence and this is a ratio of any two consecutive terms and it's called a common ratio. So here we have the definition of a geometric sequence. We have the sequence a sub 1, a sub 2, a sub 3 all the way to a sub n. k and n again are natural numbers, k again has to be less than n and we have a sub k plus 1 divided by a sub k is equal to that r for all k, all natural numbers. Then we have a sequence that is geometric and again what does this mean? Well remember we had a sub k plus 1 minus a sub k when we are doing arithmetic. So this just means take the second term and divide it by the one in front of it. So let's try. Determine if it's geometric or not, if it is find the common ratio, if it isn't try to find a pattern. So we're going to look at this and say well negative 6 divided by 3 and that's going to give us negative 2 and 12 divided by negative 6 is going to be negative 2 and negative 24 divided by 12 is going to be negative 2 and 48 divided by negative 24 is also going to be negative 2 so that we would say that our common ratio is negative 2. So that makes it geometric. So we'll determine again if it's geometric or not. So here we have negative 9 divided by negative 13 really just gives me 9 over 13. I can't really do much more with those since 13 is a prime number. And if I try the next one, negative 5 over negative 9, well again I get a positive but it's not the same ratio of 9 to 13. Because of that it doesn't matter if I were to have the same ratio for everything else beyond that because I found two consecutive numbers, pairs like that that didn't give me the same ratio then it's not geometric. But it asks us to find a pattern. Well I'm not multiplying the same thing but am I adding the same thing? It's one thing I might be able to do. Negative 13 to negative 9 maybe we added 4. Negative 9 plus 4 would be negative 5. Negative 5 plus 4 would be negative 1 and negative 1 plus 4 would be 3. So we could say that we had thinking that we started with negative 13 and we were adding 4 every time and we could think about it as being in but let's double check that that works. Put in as my first term the negative 13 plus 4 times 1 is not going to be negative 13 but if you remember with the arithmetic sequences this looks linear then we would have n minus 1 or we should have negative 13 plus 4 n minus 4. So we could say a sub n is equal to 4 n minus 17. So write the first four terms of a sequence given a sub 1 equal 2 and r is equal to negative 4. That's not so bad. A sub 1 is just going to be 2 and a sub 2 is going to be that 2 times a negative 4 and 2 times negative 4 is going to be negative 8. So far my sequence looks like it's 2 negative 8 and a sub 3 is going to be 2 times negative 4 and times negative 4. So we have that positive 32 and a sub 4 is going to be 2 times 4 times actually times negative 4 times negative 4 times negative 4 and that's going to give us 64 times 2 and it's a negative because we got an odd number of them and that's going to be a negative 128. So just like we had added number over and over and over again we're adding common multiples here. A sub 1 would be 3 times 2 to the 0 so that we still keep 3 and 3 times 2 to the 1st would give me 3 times 2 and then I need another factor so I would have 3 times 2 squared and I need one more factor here so 3 times 2 to the 3rd a sub n in general would be and again look and see what's happening here we have n is 4 and we need 3 of those n is 3 and I need 2 of my common ratio n is 2 and I need 1 of those so if I take 2 minus 1 I get 1, if I take 3 minus 1 I get 2, if I take 4 minus 1 I get 3 so this must be the number of my term minus 1 and that's a general term for a geometric sequence. If you look at this general term here you can see that that looks very exponential and if we were to actually graph it would be there are a bunch of dots because they're just terms but it would look exponential just like arithmetic looked very linear. Our common ratio is our base and the a sub 1 is our first term which we would expect to be. Alright so we want to identify the first term and the common ratio and then write the nth term so a sub 1 is going to be that very first thing we see negative 7 over 8 and then r is going to be remember you take the second term and you divide it by the first term and if you remember your fractions that means that we can multiply by the reciprocal so 7 over 4 times 8 over negative 7 that gives me 8 to 7's cancel but the negative doesn't okay so this changes to negative 1 down here and so I have 8 over negative 4 and 8 divided by negative 4 is going to be negative 2 and if I double check that just to do one more to make sure it was geometric I would have negative 7 over 2 divided by I like this 7 over 4 but remember that's the same thing as multiplying times 4 over 7 again my 7's cancel and I'm left with negative times 4 over 2 which is negative 2 so I think that really is my r negative 2 is my r a sub 1 is my negative 7 8 and remember the a sub n is equal to a sub 1 times r to the n minus 1 so we are going to have a sub n equal to negative 7 over 8 times negative 2 to the n minus 1 and then they ask us also to find a sub 6 well that's negative 7 over 8 times negative 2 to the 6 minus 1 it's really negative 7 over 8 times negative 2 to the 5th and I happen to know that that's negative 7 over 8 times negative 32 any negative times itself an odd number of times will always be negative and 8 happens to go into 32 four times so I really have negative 7 times negative 4 which is going to be a positive 28 so here was the first part and down here was the second part so we want to find the number of terms we've done this with arithmetic and it really doesn't change just have a different formula to plug into so remember a sub n is equal to a sub 1 times r to the n minus 1 plug in what we know we know a sub n a sub n is 1 over 64 and we know a sub 1 is 16 so we have 16 and we also know that r is one half so we have one half in here but I do not know what n is so I still have to keep it as n minus 1 what do you do from here you got to get to that base by itself so we need to take 1 over 64 I'm going to do it over here 1 over 64 divided by 16 which is the same thing as 1 over 64 times 1 over 16 and that happens to be 1 over 1024 but I know that I needed one half here this is 1 over 1024 is equal to one half to the n minus 1 I am going to look at this and say okay 2 to the 4th is 16 and I believe that 2 to the 6th I'll come over to my calculator double check that 2 to the 6th is equal to 64 and remember if you have the same base and you're multiplying that's going to be 2 to the 4 plus 6 or 10 so 1 over 2 to the 10 is equal to 1 over 2 to the n minus 1 and if I take them all to the reverse them I have 2 to the 10 is equal to 2 to the n minus 1 so that's telling me that 10 is equal to n minus 1 so n is equal to 11 or 11 terms now if I hadn't worked out that I had the same basis so then I would have had to take a log or do some other things but that was good find the common ratio and the first term given this information we have a sub n is equal to a sub 1 times r to the n minus 1 I'm going to take a sub n and I'm going to call the a sub 9 and I'm really going to make this a sub 1 a sub 5 and I have my r which I don't know and I'm going to say that this is if there's 5 here and I have to get to 9 that must be r to the 4th but that still doesn't do much because I got these a's in here but I know that a sub 9 is 486 and I know that a sub 5 is 6 and we can keep that in red r to the 4th and if we divide off 486 divided by 6 is going to be 81 going to be equal to r to the 4th and if I take the 4th root now I want my r to be a nice number and it is it's going to be 3 it's kind of like I'm taking the absolute value of r to the 4th because it should be plus or minus 3 but we're just going to let it be 3 I then would say that r is 3 or I found the common ratio not a sub 1 so again a sub n is equal to a sub 1 times we know r now in fact let's change that so we show that that's something that we know to the n minus 1 well if I let this be a sub 5 is equal to a sub 1 times 3 I have 1 here I need this number to add up to 5 so that would be 4 and again that's because 1 plus 4 is equal to 5 and over here it was because 5 plus 4 was equal to 9 5 plus 4 equal 9 and 1 plus 4 equal 5 plug and chug I know one more piece of information I know this is 6 is equal to a sub 1 but I'm trying to find times 3 to the 4th which happens to be 81 so a sub 1 is going to be 6 over 81 and we know that we can reduce that because they're both the visible by 3 and that would be 2 over 27 so we found the ratio and the first term