Recursive formulas for arithmetic sequences

Learn how to find recursive formulas for arithmetic sequences. For example, find the recursive formula of 3, 5, 7,...
Before taking this lesson, make sure you are familiar with the basics of arithmetic sequence formulas.

How recursive formulas work

Recursive formulas give us two pieces of information:
  1. The first term of the sequence
  2. The pattern rule to get any term from the term that comes before it
Here is a recursive formula of the sequence 3,5,7,...3, 5, 7,... along with the interpretation for each part.
{a(1)=3the first term is 3a(n)=a(n1)+2add 2 to the previous term\begin{cases}a(1) = 3&\leftarrow\gray{\text{the first term is 3}}\\\\ a(n) = a(n-1)+2&\leftarrow\gray{\text{add 2 to the previous term}} \end{cases}
In the formula, nn is any term number and a(n)a(n) is the nthn^\text{th} term. This means a(1)a(1) is the first term, and a(n1)a(n-1) is the term before the nthn^\text{th} term.
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=\greenE 3
a(2)a(2)=a(1)+2=a(1)+2=3+2=\greenE 3+2=5=\purpleC5
a(3)a(3)=a(2)+2=a(2)+2=5+2=\purpleC5+2=7=\blueD 7
a(4)a(4)=a(3)+2=a(3)+2=7+2=\blueD 7+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,...3,5,7,...

Check your understanding

1) 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)=
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}

What this formula means can be verbalized as follows:
The first term is 5-5, and any other term is the term before it plus 99.
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=\greenE{-5}
b(2)b(2)=b(1)+9=b(1)+9=5+9=\greenE{-5}+9=4=\purpleC4
b(3)b(3)=b(2)+9=b(2)+9=4+9=\purpleC4+9=13=\blueD{13}
b(4)b(4)=b(3)+9=b(3)+9=13+9=\blueD{13}+9=22=22
Therefore, b(4)=22b(4)=22.

Writing recursive formulas

Suppose we wanted to write the recursive formula of the arithmetic sequence 5,8,11,...5, 8, 11,...
The two parts of the formula should give the following information:
  • The first term ((which is 5)\greenE 5)
  • The rule to get any term from its previous term ((which is "add 3\maroonC{3}"))
Therefore, the recursive formula should look as follows:
{c(1)=5c(n)=c(n1)+3\begin{cases}c(1)=\greenE 5\\\\ c(n)=c(n-1)\maroonC{+3} \end{cases}

Check your understanding

2) What is the recursive formula of the sequence 12,7,2,...12, 7, 2,... ?
Choose 1 answer:
Choose 1 answer:

The two parts of the formula should give the following information:
  • The first term ((which is 12)\greenE{12})
  • The rule to get any term from its previous term ((which is "subtract 5\maroonC{5}"))
Therefore, the recursive formula should look as follows:
{d(1)=12d(n)=d(n1)5\begin{cases}d(1)=\greenE{12}\\\\ d(n)=d(n-1)\maroonC{-5} \end{cases}
3) Complete the missing values in the recursive formula of the sequence 2,8,14,..2,8,14,...
{e(1)=Ae(n)=e(n1)+B\begin{cases}e(1)=A\\\\ e(n)=e(n-1)+B \end{cases}
A=A=
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}
B=B=
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}

The two parts of the formula should give the following information:
  • The first term ((which is 2)\greenE{2})
  • The rule to get any term from its previous term ((which is "add 6\maroonC{6}"))
Therefore, the recursive formula should look as follows:
{e(1)=2e(n)=e(n1)+6\begin{cases}e(1)=\greenE{2}\\\\ e(n)=e(n-1)\maroonC{+6} \end{cases}
In conclusion, A=2A=2 and B=6B=6.
4) Complete the missing values in the recursive formula of the sequence 1,4,7,...-1,-4,-7,....
{f(1)=Af(n)=f(n1)+B\begin{cases}f(1)=A\\\\ f(n)=f(n-1)+B \end{cases}
A=A=
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}
B=B=
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}

The two parts of the formula should give the following information:
  • The first term ((which is 1)\greenE{-1})
  • The rule to get any term from its previous term ((which is "subtract 3\maroonC{3}"))
Therefore, the recursive formula should look as follows:
{f(1)=1f(n)=f(n1)3\begin{cases}f(1)=\greenE{-1}\\\\ f(n)=f(n-1)\maroonC{-3} \end{cases}
In conclusion, A=1A=-1 and B=3B=-3.

Reflection question

5) Here is the general recursive formula for arithmetic sequences.
{g(1)=Ag(n)=g(n1)+B\begin{cases}g(1)=A\\\\ g(n)=g(n-1)+B \end{cases}
What is the common difference of the sequence?
Choose 1 answer:
Choose 1 answer:

What this formula means can be verbalized as follows:
  • The first term is AA
  • To get any term from its previous term, add BB.
The thing that we add to each term to get the term that comes next is the common difference, so the common difference is simply BB.
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