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:
3,5,7,...3, 5, 7,...
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 nn to represent any term number and a(n)a(n) to represent the nthn^\text{th} term of the sequence. For example, here are the first few terms of the arithmetic sequence 3, 5, 7, ...
nna(n)a(n)
(The term number)(The nthn^\text{th} term)
1133
2255
3377
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(n)a(n) for any possible nn.

Check your understanding

1) Find a(4)a(4) in the sequence 3, 5, 7, ...
a(4)=a(4)=
a(4)a(4) is the fourth term of the sequence 3, 5, 7, ...
+2\footnotesize\maroonC{+2\,\Large\curvearrowright}+2\footnotesize\maroonC{+2\,\Large\curvearrowright}+2\footnotesize\maroonC{+2\,\Large\curvearrowright}
33557799
a(1)a(1)a(2)a(2)a(3)a(3)a(4)a(4)
Therefore, a(4)=9a(4)=9.
2) For any term number nn, what does a(n1)a(n-1) represent?
Choose 1 answer:
Choose 1 answer:
Let's look at a few specific values for nn:
nna(n1)a(n-1)Interpretation of a(n1)a(n-1)
2\purpleC2a(21)=a(1)a(\purpleC2-1)=a(1)The first term, which is the term before a(2)a(\purpleC2)
3\greenD3a(31)=a(2)a(\greenD3-1)=a(2)The second term, which is the term before a(3)a(\greenD3)
4\goldD4a(41)=a(3)a(\goldD4-1)=a(3)The third term, which is the term before a(4)a(\goldD4)
In general, we see that a(n1)a(n-1) is the term that comes before a(n)a(n).

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.
{a(1)=3The first term is three.a(n)=a(n1)+2Add two to the previous term.\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(n)a(n)=a(n1)+2=a(n\!-\!\!1)+2
a(1)a(1)=3=\blueD3
a(2)a(2)=a(1)+2=a(1)+2=3+2=\blueD3+2=5=\purpleC5
a(3)a(3)=a(2)+2=a(2)+2=5+2=\purpleC5+2=7=\greenD7
a(4)a(4)=a(3)+2=a(3)+2=7+2=\greenD7+2=9=\goldD9
a(5)a(5)=a(4)+2=a(4)+2=9+2=\goldD9+2=11=11
Cool! This formula gives us the same sequence as described by 3, 5, 7, ...

Check your understanding

Now it's your turn to find terms of sequences using their recursive formulas.
Just as we used a(n)a(n) to represent the nthn^\text{th} term of the sequence 3, 5, 7, ..., we can use other letters to represent other sequences. For example, we can use b(n)b(n), c(n)c(n), or d(n)d(n).
3) Find b(4)b(4) in the sequence given by {b(1)=5b(n)=b(n1)+9\begin{cases}b(1)=-5\\\\ b(n)=b(n-1)+9 \end{cases}
b(4)=b(4)=
The meaning of this formula can be verbalized as follows:
The first term is negative five, and any other term is the term before it plus nine.
In order to find b(4)b(4), we need to extend the sequence term by term:
b(n)b(n)=b(n1)+9=b(n\!-\!\!1)+9
b(1)b(1)=5=\blueD{-5}
b(2)b(2)=b(1)+9=b(1)+9=5+9=\blueD{-5}+9=4=\purpleC4
b(3)b(3)=b(2)+9=b(2)+9=4+9=\purpleC4+9=13=\greenD{13}
b(4)b(4)=b(3)+9=b(3)+9=13+9=\greenD{13}+9=22=22
Therefore, b(4)=22b(4)=22.
4) Find c(3)c(3) in the sequence given by {c(1)=20c(n)=c(n1)17\begin{cases}c(1)=20\\\\ c(n)=c(n-1)-17 \end{cases}
c(3)=c(3)=
The meaning of this formula can be verbalized as follows:
The first term is 20, and any other term is the term before it minus 17.
In order to find c(3)c(3), we need to extend the sequence term by term:
c(n)c(n)=c(n1)17=c(n\!-\!\!1)-17
c(1)c(1)=20=\blueD{20}
c(2)c(2)=c(1)17=c(1)-17=2017=\blueD{20}-17=3=\purpleC3
c(3)c(3)=c(2)17=c(2)-17=317=\purpleC3-17=14=-14
Therefore, c(3)=14c(3)=-14.
5) Find d(5)d(5) in the sequence given by {d(1)=2d(n)=d(n1)+0.4\begin{cases}d(1)=2\\\\ d(n)=d(n-1)+0.4 \end{cases}
d(5)=d(5)=
The meaning of this formula can be verbalized as follows:
The first term is two, and any other term is the term before it plus 0.4.
In order to find d(5)d(5), we need to extend the sequence term by term:
d(n)d(n)=d(n1)+0.4=d(n\!-\!\!1)+0.4
d(1)d(1)=2=\blueD{2}
d(2)d(2)=d(1)+0.4=d(1)+0.4=2+0.4=\blueD{2}+0.4=2.4=\purpleC{2.4}
d(3)d(3)=d(2)+0.4=d(2)+0.4=2.4+0.4=\purpleC{2.4}+0.4=2.8=\greenD{2.8}
d(4)d(4)=d(3)+0.4=d(3)+0.4=2.8+0.4=\greenD{2.8}+0.4=3.2=\goldD{3.2}
d(5)d(5)=d(4)+0.4=d(4)+0.4=3.2+0.4=\goldD{3.2}+0.4=3.6=3.6
Therefore, d(5)=3.6d(5)=3.6.

Explicit formulas of arithmetic sequences

Here is an explicit formula of 3, 5, 7, ...
a(n)=3+2(n1)a(n)=3+2(n-1)
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=5n=5 into the explicit formula.
a(5)=3+2(51)=3+24=3+8=11\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!

Check your understanding

6) Find b(10)b(10) in the sequence given by b(n)=5+9(n1)b(n) = -5+9(n-1).
b(10)=b(10)=
b(10)=5+9(101)=5+99=5+81=76\begin{aligned}b(\greenE{10})&=-5+9(\greenE{10}-1)\\\\ &=-5+9\cdot9\\\\ &=-5+81\\\\ &=76\end{aligned}
7) Find c(8)c(8) in the sequence given by c(n)=2017(n1)c(n) = 20-17(n-1).
c(8)=c(8)=
c(8)=2017(81)=20177=20119=99\begin{aligned}c(\greenE{8})&=20-17(\greenE{8}-1)\\\\ &=20-17\cdot7\\\\ &=20-119\\\\ &=-99\end{aligned}
8) Find d(21)d(21) in the sequence given by d(n)=2+0.4(n1)d(n) = 2+0.4(n-1).
d(21)=d(21)=
d(21)=2+0.4(211)=2+0.420=2+8=10\begin{aligned}d(\greenE{21})&=2+0.4(\greenE{21}-1)\\\\ &=2+0.4\cdot20\\\\ &=2+8\\\\ &=10\end{aligned}

Sequences are functions

Notice that the formulas we used in this lesson work like functions: We input a term number nn, and the formula outputs the value of that term a(n)a(n).
Sequences are in fact defined as functions. However, nn 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.

A note about notation

We've been writing a(4)a(4), for example, to represent the fourth term, but other sources sometimes write a4a_4.
Both notations are fine to use. We prefer a(4)a(4) 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?
Choose 1 answer:
Choose 1 answer:
To find the 100th term using the recursive formula, we have to extend the sequence term-by-term until we get to 100. That will take a very long time!
To find the 100th term using the explicit formula, all we need to do is to plug n=100n=100 into the formula and evaluate. This doesn't take more time than finding the second term, or the 1000th term, for that matter.
In conclusion, the explicit formula is more useful for finding the 100th term of an arithmetic sequence.

Challenge problem

10) The explicit formula of an arithmetic sequence is f(n)=34(n1)f(n) =3-4(n-1).
Which term of the sequence is equal to -65?
Term number
.
The number of the term is represented in the formula by nn, and its value is represented by f(n)f(n).
In this case, we know the value of f(n)f(n), and we want to find nn. This calls for solving an equation.
f(n)=34(n1)65=34(n1)Substitute f(n)=65.4(n1)=68n1=17n=18\begin{aligned}f(n)&=3-4(n-1)\\\\ -65&=3-4(n-1)&\gray{\text{Substitute }f(n)=-65.}\\\\ 4(n-1)&=68\\\\ n-1&=17\\\\ n&=18\end{aligned}
In conclusion, the term that is equal to -65 is term number 18.
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