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 plus thirteen times the where the know what the is right,
these negative nine how do we know that these negative nine all that stands we just got
here okay so equals ninety-one sixteen solve that we have A1 plus, negative nine times
of thirteen is negative one seventeen changes here today have A1 equals negative one seventeen
equals one negative one sixteen in a get the one by itself by adding one seventeen to both
sides the have A1 equals positive one right so common difference whether common difference
our common difference the is negative and I write so that basically means that every
time will going to subtract.
To generate terms of our sequence is sick at the be part to be parts we are to find
the fifty first to find of a fifty-one this is find the fifty first are so what follows
going to use all the same end term formula a and equals A1 plus and minus one times the
A1 is one that's the answer we got ina and is the index of the term we looking for which
is fifty-one and then our common difference the con different
D is the negative nine of a.m.'s and is equal to a fifty-one and that's what we looking
for okay so let's compute what the Valley of a fifty one is using of the sum this of
information that we've assembled so we going to have of the going to have a fifty-one is
equal to A1 which is one plus and fifty-one minus one times the common difference which
is negative nine right's go ahead and simplify this we going to have one plus fifty times
negative nine the order of operations we're going to have one minus four hundred and fifty
which is equal to negative four right cursor four forty-nine is for that again negative
four forty-nine right so that's what our fifty first term is negative four forty-nine's right
so, going to the next parts we are to find the explicit formula of and then verify that
our results in part B is in fact correct.
So for the explicit formula of him will going to use the same formula that we use actually
have only one choice is the of formula right to find X is it formula of and is going to
be equal to ten right that's the difference and is equal to and A1 is always going to
be one in a common difference is always going to be negative nine so these are the of to
the of information with an up at together to generate our explicit formula for the end
term of our arithmetic sequence with the given terms okay so substituting the going to have
a and is equal to of A1 which is one plus and which is to and one is one times the common
difference which is negative nine right so just of go ahead and simplify this formula
related have a and is equal to one now there distribute this negative nine to these two
terms in the parenthesis so we have negative nine and plus nine simplify further we have
a and equals negative nine and plus ten right so this is our explicit formula explicit formula
now of is it correct over going to do is will going to of verify the accuracy of this result
by of determining by finding the fifty first term using this formula in and see if we have
the same answer as part B okay so verification verification will going to test it okay with
an equals fifty-one are we going to get negative four forty-nine that's the question.
So let's would testing our results in part C a fifty one is equal to negative nine times
fifty-one plus ten okay so multiply this year going to have negative four fifty-nine plus
ten which equals negative four forty-nine that's perfect for checks out because that's
what we had here okay so if our answer answer our X is it formula is correct right is move
on to depart we are to find the recursive formula recursive formula while of recursive
formulas pretty straightforward to come up with a and equals now we ask yourself in this
in this of sequence what do you do to generate subsequent terms you subtract nine every time
okay so from the term before you subtract right to get the next term so if and is the
next term what you do to the term before that's of one or I this is a list you have a and
write what is the term before a and and the sequence determine for in the sequence is
a and minus one okay so what you do to term before to get the next and what is the today
and to get twenty and to what he do to the and minus one to get and you subtract nine
that's how the pattern is you subtract right every time so you the recursive definition
a we just simply right is the next term and can be found by taking the term before it
and doing what subtract and nine simply shirts that's what the recursive definitions of lastly
we are to find the first of four terms of the sequence for support terms of so is going
to find that first four terms we have the first term is one the second term eight two
remember what the recursive definition is you we subtract one right so the term before
answer your we subtract nine so the term before A1 minus nine gives us negative eight sign
X two eight three is going to be the term before minus nine again which is negative
seventeen and then the fourth term is going to be a term before minus nine which is negative
twenty-six right so first four times our first four terms are of one negative eight negative
seventeen and negative twenty-six right so that's that the thanks a much for taking the
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