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# Intro to arithmetic sequence formulas

Get comfortable with the basics of explicit and recursive formulas for arithmetic sequences.
Before taking this lesson, make sure you know the basics of arithmetic sequences and have some experience with evaluating functions and function domain.

## What is a formula?

We are used to describing arithmetic sequences like this:
3, comma, 5, comma, 7, comma, point, point, point
But there are other ways. In this lesson, we'll be learning two new ways to represent arithmetic sequences: recursive formulas and explicit formulas. Formulas give us instructions on how to find any term of a sequence.
To remain general, formulas use n to represent any term number and a, left parenthesis, n, right parenthesis to represent the n, start superscript, start text, t, h, end text, end superscript term of the sequence. For example, here are the first few terms of the arithmetic sequence 3, 5, 7, ...
na, left parenthesis, n, right parenthesis
(The term number)(The n, start superscript, start text, t, h, end text, end superscript term)
13
25
37
We mentioned above that formulas give us instructions on how to find any term of a sequence. Now we can rephrase this as follows: formulas tell us how to find a, left parenthesis, n, right parenthesis for any possible n.

1) Find a, left parenthesis, 4, right parenthesis in the sequence 3, 5, 7, ...
a, left parenthesis, 4, right parenthesis, equals

2) For any term number n, what does a, left parenthesis, n, minus, 1, right parenthesis represent?

## Recursive formulas of arithmetic sequences

Recursive formulas give us two pieces of information:
1. The first term of a sequence
2. The pattern rule to get any term in a sequence from the term that comes before it
Here is the recursive formula of our sequence 3, 5, 7, ... along with the interpretation for each part.
$\begin{cases}a(1) = 3&\leftarrow\gray{\text{The first term is three.}}\\\\ a(n) = a(n-1)+2&\leftarrow\gray{\text{Add two to the previous term.}} \end{cases}$
In order to find the fifth term, for example, we need to extend the sequence term by term:
a, left parenthesis, n, right parenthesisequals, a, left parenthesis, n, minus, 1, right parenthesis, plus, 2
a, left parenthesis, 1, right parenthesisequals, start color #11accd, 3, end color #11accd
a, left parenthesis, 2, right parenthesisequals, a, left parenthesis, 1, right parenthesis, plus, 2equals, start color #11accd, 3, end color #11accd, plus, 2equals, start color #aa87ff, 5, end color #aa87ff
a, left parenthesis, 3, right parenthesisequals, a, left parenthesis, 2, right parenthesis, plus, 2equals, start color #aa87ff, 5, end color #aa87ff, plus, 2equals, start color #1fab54, 7, end color #1fab54
a, left parenthesis, 4, right parenthesisequals, a, left parenthesis, 3, right parenthesis, plus, 2equals, start color #1fab54, 7, end color #1fab54, plus, 2equals, start color #e07d10, 9, end color #e07d10
a, left parenthesis, 5, right parenthesisequals, a, left parenthesis, 4, right parenthesis, plus, 2equals, start color #e07d10, 9, end color #e07d10, plus, 2equals, 11

Cool! This formula gives us the same sequence as described by 3, 5, 7, ...

Now it's your turn to find terms of sequences using their recursive formulas.
Just as we used a, left parenthesis, n, right parenthesis to represent the n, start superscript, start text, t, h, end text, end superscript term of the sequence 3, 5, 7, ..., we can use other letters to represent other sequences. For example, we can use b, left parenthesis, n, right parenthesis, c, left parenthesis, n, right parenthesis, or d, left parenthesis, n, right parenthesis.
3) Find b, left parenthesis, 4, right parenthesis in the sequence given by $\begin{cases}b(1)=-5\\\\ b(n)=b(n-1)+9 \end{cases}$
b, left parenthesis, 4, right parenthesis, equals

4) Find c, left parenthesis, 3, right parenthesis in the sequence given by $\begin{cases}c(1)=20\\\\ c(n)=c(n-1)-17 \end{cases}$
c, left parenthesis, 3, right parenthesis, equals

5) Find d, left parenthesis, 5, right parenthesis in the sequence given by $\begin{cases}d(1)=2\\\\ d(n)=d(n-1)+0.4 \end{cases}$
d, left parenthesis, 5, right parenthesis, equals

## Explicit formulas of arithmetic sequences

Here is an explicit formula of 3, 5, 7, ...
a, left parenthesis, n, right parenthesis, equals, 3, plus, 2, left parenthesis, n, minus, 1, right parenthesis
This formula allows us to simply plug in the number of the term we are interested in to get the value of that term.
In order to find the fifth term, for example, we need to plug n, equals, 5 into the explicit formula.
\begin{aligned}a(\greenE 5)&=3+2(\greenE 5-1)\\\\ &=3+2\cdot4\\\\ &=3+8\\\\ &=11\end{aligned}
Lo and behold, we get the same result as before!

6) Find b, left parenthesis, 10, right parenthesis in the sequence given by b, left parenthesis, n, right parenthesis, equals, minus, 5, plus, 9, left parenthesis, n, minus, 1, right parenthesis.
b, left parenthesis, 10, right parenthesis, equals

7) Find c, left parenthesis, 8, right parenthesis in the sequence given by c, left parenthesis, n, right parenthesis, equals, 20, minus, 17, left parenthesis, n, minus, 1, right parenthesis.
c, left parenthesis, 8, right parenthesis, equals

8) Find d, left parenthesis, 21, right parenthesis in the sequence given by d, left parenthesis, n, right parenthesis, equals, 2, plus, 0, point, 4, left parenthesis, n, minus, 1, right parenthesis.
d, left parenthesis, 21, right parenthesis, equals

## Sequences are functions

Notice that the formulas we used in this lesson work like functions: We input a term number n, and the formula outputs the value of that term a, left parenthesis, n, right parenthesis.
Sequences are in fact defined as functions. However, n cannot be any real number value. There's no such thing as the negative fifth term or the 0.4th term of a sequence.
This means that the domain of sequences—which is the set of all possible inputs of the function—is the positive integers.

We've been writing a, left parenthesis, 4, right parenthesis, for example, to represent the fourth term, but other sources sometimes write a, start subscript, 4, end subscript.
Both notations are fine to use. We prefer a, left parenthesis, 4, right parenthesis because it emphasizes that sequences are functions.

### Reflection question

9) Which type of formula is more useful for quickly finding the 100th term of an arithmetic sequence?

### Challenge problem

10) The explicit formula of an arithmetic sequence is f, left parenthesis, n, right parenthesis, equals, 3, minus, 4, left parenthesis, n, minus, 1, right parenthesis.
Which term of the sequence is equal to -65?
Term number
.

## Want to join the conversation?

• If explicit formulas of the sequence are easier to find the 100th term of a sequence (based on the reflection question), why would we need recursive formulas? When are they useful? • I have a question about your opinion on notation. So I am currently teaching this concept and after seeing this, I feel my students would have felt so much more comfortable using the notation you have as representing the arithmetic sequence as a function. The book we use uses asubn notation. Do you think there is a reason to use one notation over the other? • I feel like a robot using the explicit formula. Why on earth does it work? It seems phony, since we are always given the formulas that define the sequence in each exercise. How does one arrive to such formula? • what is difference between explicit formula and recursiue formula because both equation looks similar jus we changed the side of our common difference d.
i mean a(n)=a+(n-1)D or a(n)=a+D(n-1)
we just changed the side of D .
SO are they similar? • Actually the explicit formula for an arithmetic sequence is a(n)=a+(n-1)*D, and the recursive formula is a(n) = a(n-1) + D (instead of a(n)=a+D(n-1)).

The difference is than an explicit formula gives the nth term of the sequence as a function of n alone, whereas a recursive formula gives the nth term of a sequence as a function of the previous term(s) of the sequence (and sometimes of n as well). In other words, an explicit formula tells how to calculate directly the nth term of a sequence, whereas a recursive formula tells how to get from one or more terms of the sequence to the next term.

Have a blessed, wonderful day!
• In problem seven it gives the problem Find c(8) in the sequence given by c(n)= 20-17(n-1) okay so how would i solve that problem i thought that you multiplied 17 by 17 and then subtracted it by 20 but when you click on the help button it says
c(8)=20-17 (8-1)
=20-17*7
=20-119
= -99
So my question is how they got the 17*7 and not 17*17 • How are recursive formulas still functions? The second part of the recursive formula seems to me like using a word in that word's definition (because the letter used to show the function is used in the definition of the function, if that makes sense...?). • A function specifies a relationship between two variables. While a recursive formula does use itself to sequence back to its initial value, the result is still specifying the relationship between two variables.

Also notice that one can change arithmetic sequences to recursive formulas and vice versa. Both are functions.

Recursion itself, whether in math or in other places such as computer programming, can be a very different way of looking at a problem. Recursion is not limited to mathematics, but that's a whole different topic all together.
• Is a(n) more confusing than "a" sub "n"?
It makes me think "a multiplied to n" sometimes. • I still don't understand what you mean. I tried clicking 'I need help' but that didn't explain it well enough for me to understand • The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P. • assume that the common difference is y and the first term is x, so
x + x + y + x + 2y + x + 3y + x + 4y + x + 5y + x + 6y = 63 which means 7x + 21y = 63
x + 7y + x + 8y + x + 9y + x + 10y + x + 11y + x + 12y + x + 13y = 161 so 7x + 70y = 161, subtract the first equation from the second and get 49y = 98, so y =2 and substitute this into either equation,
7x + 21(2) = 63, 7x = 21, x = 3. x is the first term, and two is the common difference, so f(n) = 2(n-1) + 3. f(28)= 2(27) + 3 = 57. 