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### Course: Algebra 1 > Unit 9

Lesson 2: Constructing arithmetic sequences- Recursive formulas for arithmetic sequences
- Recursive formulas for arithmetic sequences
- Recursive formulas for arithmetic sequences
- Explicit formulas for arithmetic sequences
- Explicit formulas for arithmetic sequences
- Explicit formulas for arithmetic sequences
- Arithmetic sequence problem
- Converting recursive & explicit forms of arithmetic sequences
- Converting recursive & explicit forms of arithmetic sequences
- Converting recursive & explicit forms of arithmetic sequences
- Arithmetic sequences review

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# Explicit formulas for arithmetic sequences

Sal finds explicit formulas of arithmetic sequences given the first few terms of those sequences. He also explores equivalent forms of such formulas.

## Want to join the conversation?

- I'm a little confused. So it states f(n)=12-7(n-1), so if n=4 we have

f(n)= 5(3) = 15 this is wrong tho.....

But if n=2 we get f(n)=12-7(2-1) =5 which is correct.(10 votes)- Just use Order of Operations, and you will get the right answer for every term

So for n=4, first use the equation f(n) = 12 - 7(n - 1), plug in 4 for n. Then, in the parenthesis, you will have 4-1, which is 3. Then, multiply 7*3 = 21. Lastly, subtract 12 from 21, to get -9, which is the correct answer. When using arithmetic sequence formula. Always do the operation inside the parenthesis first, then multiply the result by the number outside the parenthesis( this is the common difference). Lastly take the product of that operation, and subtract/add (depends on the product!) to the first number ( which is the first term of the sequence. Do this, and because you are using order of operations, you will find the right term, no matter what sequence it is.(12 votes)

- who is the guy who makes all these videos?(2 votes)
- The "guy" is Salman "Sal" Khan, the founder of the site.

Go to this page: https://www.khanacademy.org/about/the-team for a short bio:

"Sal

Founder & CEO

Sal started Khan Academy in 2005 to help his cousins (and soon other people's cousins). In addition to setting the vision and direction for Khan Academy, he still makes a lot of videos (although he's not the only one anymore).

Sal holds three degrees from MIT and an MBA from Harvard."

Or here: https://en.wikipedia.org/wiki/Sal_Khan for a longer one.(34 votes)

- At about2:02Sal shows the explicit formula, but in school I learned it the way shown below.

Is the formula I use and the formula in this video the same?

12, 5, -2, -9

a1=-7n+19

a1=-7(1)+19

a1=-7+19

a1=12

a2=-7n + 19

a2=-7(2)+19

a2=-14+19

a2=5

a3=-7n + 19

a3=-7(3)+19

a3=-21+19

a3=-2

a4=-7n + 19

a4=-7(4)+19

a4=-28+19

a4=-9(4 votes)- Your function is ok. If you take Sal's function and simplify it, then you get your version. Sal's version is a little more common form because it can quickly be converted to recursive form.

Hope this helps.(5 votes)

- But what if you have to find a sequence in between two other sequences? How would you solve it then? Is there another video a problem like that?(5 votes)
- Is f(n) = 12 - 7 (n - 1) same as f(n) = 12 - 3.5 (n - 2)?

Plz help

Thanks!(2 votes)- You can determine this by trying some value of "n".

If n=1:

Your first equation creates: 12-7(1-1) = 12-0 = 12

Your second equation creates: 12-3.5(1-2) = 12+3.5 = 15.5

So, they aren't the same. They create different output values.(6 votes)

- Okay if I wanted to know the arithmetic sequence how do I go about solving for a new arithmetic sequence from a previous arithmetic sequence, how do I go about it, here is my example:

17, 22, 36, 37, 52, 24 ----to get----> = 14, 22, 52, 54, 59, 4

20, 21, 23, 38, 42, 6 -----to get---> = 22, 27, 35, 37, 45, 3

especially having to handle the general term of "n" as part of this problem. How would I go about finding a sequential solution to this problem?(4 votes)- How to solve this problem : for a particular sequence the first term is 3 and the explicit formula is t sub n = -2tsub n -1 + 1, find the sixth term(2 votes)

- When I was a teen this stuff wasn't talked about at all in school; I got all the way through two college pre-calc classes without ever seeing anything to do with sequences. I wonder what caused it to be added in?(3 votes)
- Half of school is ultimately pointless. We will never need to know how to find the sixth term in an arithmetic sequence or how to define appropriate quantities for modeling. In my opinion, regular school should end at around 5th or 6th grade, and then ages 11-18 should be spent learning real world problems (i.e. how to earn passive income, how to buy real estate, how to drive) and reviewing what they already learned in school.(3 votes)

- Probably a basic question, but is there a lesson or anything that explains why we don't distribute the -7 in f(n)=12-7(n-1)?

I'm assuming it's kind of just implied when dealing with functions, but I can't help but get the urge to distribute whenever I see a number to the left of things in parenthesis.(2 votes)- As an explicit function you totally could, when learning about arithmetic sequences you normally won't just to be super clear it was an arithmetic sequence. But the point of turning it into an explicit function is that you ca treat it like a normal linear equation.

If you have them in the recursive form when you see it in a problem it will be obvious the n-1 and whatnot won't get numbers distributed because they will be subscripts (small numbers in the lower right) but it may be a little tricky if you type it out. Just be aware if it is in recusrive form or function form.

Let me know if that didn't help.(3 votes)

- I'm having problems with this question, could anyone help me? Write the explicit formula for the arithmetic sequence.

2, -2, -6, -10, -14, ...(2 votes)- So, the common difference is -4 and the first sequence is 2. Therefore, the explicit formula would be an = 2 - 4(n-1).

Hope this helps! If you have any questions or need help, please ask! :)(2 votes)

- How do you translate between recursive and explicit rules for arithmetic sequences?(2 votes)
- Find the first term a_1 and the common difference d. Then you can get the other formula quite easily.

Ex.) Find the recursive formula if the explicit formula is a_n = 5n - 3.

Let's find a_1 and d.

a_1 = 5 * 1 - 3 = 2

d = a_(n + 1) - a_n = 5(n + 1) - 3 - 5n + 3 = 5

Recursive formula:

a_1 = 2

a_n = a_(n - 1) + 5, where n = 2, 3, 4,...(1 vote)

## Video transcript

So we see here in this table, that for given "n's", when "n" is one, "f" of "n" is 12, when
"n" is 2, "f" of "n" is 5, and when "n" is three, "f"
of "n" is negative two, when "n" is four, "f"
of "n" is negative nine. And so one way to think about it is this function "f" is defining a sequence where the first term
of this sequence is 12. The second term of this sequence is five. The third term of this
sequence is negative two. The fourth term of the
sequence is negative nine. And it goes on, and on, and on. And you might notice that
it's a arithmetic sequence. We start with a 12, and
then the next term... What have we done? We've subtracted seven. Now to go from the second
to the third term... What do we do? We subtract seven. So each term is seven less
than the term before it. Now with that out of the
way, see if you can define this function of "n." If you
can define it explicitly. So figure out a function definition. So I want to figure out
"f" of "n" is equal. I want you to figure out
what this needs to be so that if you input "n" here, it gives you the appropriate "f" of "n." So let's think about it a little bit. It's going to be, we could think
of it as we're starting at, The first term is going to be 12. But then, we are going to subtract seven. And what are we going to subtract seven? How many times are we
going to subtract seven? So for the first term, we
subtract seven zero times. And so we just get 12. For the second term,
we subtract seven once. For the third term, we subtract
seven twice. One, two times. For the fourth term, we
subtract seven three times. So it looks like whatever term we're on, we're subtracting seven "n" minus one, we're subtracting seven
whatever term we're on, that term minus one time. So it's "n" minus one times. And let's see if this actually works out. So "f" of one is going to
be 12 minus seven times one minus one, that's a zero. So that's all just going to be 12. "f" of two is going to be 12 minus seven times two minus one. So it's going to be 12
minus seven times one. We're just going to subtract seven once, which is exactly the case. We started at 12, we subtract seven once. "f" of three, you can keep testing this. 12 minus, and we should have
to subtract seven twice. And we see three minus one, is two times. So we're going to
subtract seven two times. So this looks right on. We've defined the function explicitly. We've defined "f" explicitly
for this sequence. Let's do another example, here. So in this case, we have some function definitions already here. So you have your sequence, it's kind of viewed in this table. You could view it as the
first term is negative 100. The next term is negative
50, next term is zero, next term is 50. And it's very clear that this is also an arithmetic sequence. We're starting at negative
100, and then, we're adding 50. And then we're adding 50, and then we are adding 50. So each term is 50 more
than the term before it. And what I want you to
do is pause the video, and think about which
of these definitions of the function "f" are correct. And it might be more than one. Alright, so let's think about it. So this definition right over here, one way to think about it is by saying I'm going to start at negative 100. And I'm going to add
50, "n" minus one times. Does this make sense? Well, for the first term,
if we start at negative 100, we don't want to add 50 at all, we want to add 50 zero
times, and it works out. Because one minus one is going to be zero. So it checks out for "n" equals one. Let's see, for "n" equals two,
you start at negative 100, I want to add 50 once. So this should be a one. Two minus one, yup, it's a one! We're adding 50. Whatever this number is, whatever "n" is, we're adding 50 one less that number of times. So for here, we're adding 50 twice. When "n" is four, we're
adding 50 three times. And this one checks out. When "n" is four, we're adding 50, four minus one, three times. Negative 100, plus 50 times three. We're adding 50 three times, adding 50 one, two, three times. Well that gives you
50. So I like this one. Now let's see about this one over here. Negative 150 plus 50 "n." Alright, that's one way
of saying, so let's see If "n" is equal to one,
it's going to be negative... Actually, let me draw
a table for this one. So if we have "n" and we have "f" of "n." This is going to be for this
character right over here. So if "n" is one, it's going
to negative 150, plus 50. Which is negative 100,
yup that checks out. When "n" is two, we get negative 150, plus 50 times two, which is going to be... This is 100, and there's negative 150, this is going to be negative 50. When "n" is three, and
that checks out, of course. When "n" is three, you get negative 150, plus 50, times three,
which is equal to zero. This checks out. This one over here is going to work. And you might say, "Well, hey, these formulas look different." Well you can algebraically manipulate them to see that they're the exact same thing. If you were to take this first one, it's negative 100, plus,
let's distribute this 50, plus 50 "n", plus 50 "n", minus 50. Well, negative 100, minus
50, that's negative 150. And then you have plus 50 "n." So these are algebraically the exact same definition for our function. Now what about this one here? Negative 100 plus 50
"n", does this one work? Let's see, when "n" is equal to one, this would be negative 100
plus 50, which is negative 50. Well no, this doesn't work. We need to get a negative 100 here. So this one is not, not correct.