Day students in this clip were going to be going over arithmetic sequences of the relearning how to find a common difference end term explicit and recursive formulas right so let's write down the of instructions for the example of we're. It's as follows for the given two terms can two terms of an arithmetic sequence to do it arithmetic sequences here of an arithmetic sequence of we are to find the following a find the common difference, and difference of be Valley to find the fifth the first term, find the fifty first term in a we are to of find the explicit formula X is it formula a we're not just going to find X is it formula going to use it to verify the result we got in part the occasion and verify the result in part D okay, and in part D we are to of find the recursive formula we have the explicit formula in part C in part D the going to find the recursive formula and the lastly, but to find of the of first four terms first four terms of the of the sequence is of the sequence right so the of two terms that are provided for the arithmetic sequence under consideration are a fourteen which is equal to negative one hundred and sixteen and a thirtieth which is equal to negative two hundred and sixty right so let's go ahead and of answer these questions part a one of find a common difference now what were working with arithmetic sequences there is only one formula that we have to work with which is the formula the end term formula a and equals A1 the first term plus and minus one times the common difference the so these two terms give us some valuable information that can enable us to set up on some equation so first one a fourteen what does that mean means that the this is the fourteenth term so for a fourteen and is equal to fourteen okay now is an is equal to fourteen then we can substitute the values a of fourteen and and into this equation right here so says and it is an is equal to fourteen of follows that and the negative sixty-one sixteen is equal to A1 we do not know what the first term is a one of plus and which is fourteen minus one times the common difference. The brothers look at what we just did again in this case when and is equal to fourteen a.m. is going to be negative one sixteen A1 see the same and is fourteen minus one D right and if we simplify this this becomes and what we switch it around A1 plus fourteen minus one is thirteen thirteen D equals negative one sixteen now how about this term here and is equal to thirty can you generate the formula an equation using this formula lamplight is equal to thirty and eight thirty is equal to negative sixty-two this, right here with this I just like we did for and equals fourteen a have negative to sixteen equals A1 the first term plus of and in this case is thirty minus one times the common difference the right so let's simplify this if we switch it around the have A1 plus thirty minus one is twenty-nine D equals the thirty X term which is negative two hundred and sixty right now what do we have here this is a system of equations the system of linear equations and consult this by elimination her substitution spell assaulted by lamination the can eliminate unit A1 is in Salford D or eliminated these are so for anyone of which one is easy to accomplish here since it's easier to get the A1 is to the opposites of each other out. We can eliminate that one. So let's eliminate A1 okay to do that out this multiply the second equation by negative one which makes this the opposite of that so we combine we can eliminate A1 right so for do that we have the first equation remain unchanged A1 plus thirteen D equals negative one sixteen's and then the second equation becomes negative A1 minus twenty-nine D equals two hundred and sixty right is go ahead and add downwards and we add downwards will gets its negative sixteen D equals one hundred and forty-four to get the isolated will divide both sides by negative sixteen and we of our final result to be D equals negative nine Right now we have to have the answer for our part a the common difference is negative.
But so let's go ahead and find A1 while added so that on we can use to answer the other questions rights so to find A1 of two to find A1 we just simply going to substitute the Valley of the into one of these two equations to so for of A1 okay so this is the first one so that I you can go A1
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