Converting recursive & explicit forms of arithmetic sequences

CCSS Math: HSF.BF.A.2
Learn how to convert between recursive and explicit formulas of arithmetic sequences.
Before taking this lesson, make sure you know how to find recursive formulas and explicit formulas of arithmetic sequences.

Converting from a recursive formula to an explicit formula

An arithmetic sequence has the following recursive formula.
{a(1)=3a(n)=a(n1)+2\begin{cases} a(1)=\greenE 3 \\\\ a(n)=a(n-1)\maroonC{+2} \end{cases}
Recall that this formula gives us the following two pieces of information:
  • The first term is 3\greenE 3
  • To get any term from its previous term, add 2\maroonC 2. In other words, the common difference is 2\maroonC 2.
Let's find an explicit formula for the sequence.
Remember that we can represent a sequence whose first term is A\greenE A and common difference is B\maroonC B with the standard explicit form A+B(n1)\greenE A+\maroonC B(n-1).
Therefore, an explicit formula of the sequence is a(n)=3+2(n1)a(n)=\greenE{3}\maroonC{+2}(n-1).

Check your understanding

1) Write an explicit formula for the sequence.
{b(1)=22b(n)=b(n1)+7\begin{cases} b(1)=-22 \\\\ b(n)=b(n-1)+7 \end{cases}
b(n)=b(n)=

The standard explicit form for arithmetic sequences is A+B(n1)\greenE A+\maroonC B(n-1) where A\greenE A is the first term and that B\maroonC B is the common difference.
According to the recursive formula,
  • the first term is 22\greenE{-22}, and
  • the common difference is 7\maroonC{7}.
In conclusion, b(n)=22+7(n1)b(n)=\greenE{-22}\maroonC{+7}(n-1).
2) Write an explicit formula for the sequence.
{c(1)=8c(n)=c(n1)13\begin{cases} c(1)=8 \\\\ c(n)=c(n-1)-13 \end{cases}
c(n)=c(n)=

The standard explicit form for arithmetic sequences is A+B(n1)\greenE A+\maroonC B(n-1) where A\greenE A is the first term and that B\maroonC B is the common difference.
According to the recursive formula,
  • the first term is 8\greenE{8}, and
  • the common difference is 13\maroonC{-13}.
In conclusion, c(n)=813(n1)c(n)=\greenE{8}\maroonC{-13}(n-1).

Converting from an explicit formula to a recursive formula

Example 1: Formula is given in standard form

We are given the following explicit formula of an arithmetic sequence.
d(n)=5+16(n1)d(n)=\greenE 5\maroonC{+16}(n-1)
This formula is given in the standard explicit form A+B(n1)\greenE A+ \maroonC B(n-1) where A\greenE A is the first term and that B\maroonC B is the common difference. Therefore,
  • the first term of the sequence is 5\greenE 5, and
  • the common difference is 16\maroonC{16}.
Let's find a recursive formula for the sequence. Recall that the recursive formula gives us two pieces of information:
  1. The first term ((which we know is 5)\greenE 5)
  2. The pattern rule to get any term from the term that comes before it ((which we know is "add 16\maroonC{16}"))
Therefore, this is a recursive formula for the sequence.
{d(1)=5d(n)=d(n1)+16\begin{cases} d(1)=\greenE 5\\\\ d(n)=d(n-1)\maroonC{+16} \end{cases}

Example 2: Formula is given in simplified form

We are given the following explicit formula of an arithmetic sequence.
e(n)=10+2ne(n)=10+2n
Note that this formula is not given in the standard explicit form A+B(n1)\greenE A+ \maroonC B(n-1).
For this reason, we can't simply use the structure of the formula to find the first term and the common difference. Instead, we can find the first two terms:
  • e(1)=10+21=12e(\blueD 1)=10+2\cdot\blueD 1=12
  • e(2)=10+22=14e(\blueD 2)=10+2\cdot\blueD 2=14
Now we can see that the first term is 12\greenE{12} and the common difference is 2\maroonC{2}.
Therefore, this is a recursive formula for the sequence.
{e(1)=12e(n)=e(n1)+2\begin{cases} e(1)=\greenE{12}\\\\ e(n)=e(n-1)\maroonC{+2} \end{cases}

Check your understanding

3) The explicit formula of an arithmetic sequence is f(n)=5+12(n1)f(n)=5+12(n-1).
Complete the missing values in the recursive formula of the sequence.
{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}

In the given recursive formula, AA is the first term and BB is the common difference.
The explicit formula is given in the standard form A+B(n1)\greenE A+ \maroonC B(n-1) where A\greenE A is the first term and that B\maroonC B is the common difference.
Since the explicit formula is 5+12(n1)\greenE{5}\maroonC{+12}(n-1), we know that the first term is 5\greenE{5} and the common difference is 12\maroonC{12}.
In conclusion, A=5\greenE A=\greenE{5} and B=12\maroonC B=\maroonC{12}.
4) The explicit formula of an arithmetic sequence is g(n)=118(n1)g(n)=-11-8(n-1).
Complete the missing values in the recursive formula of the sequence.
{g(1)=Ag(n)=g(n1)+B\begin{cases} g(1)=A\\\\ g(n)=g(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}

In the given recursive formula, AA is the first term and BB is the common difference.
The explicit formula is given in the standard form A+B(n1)\greenE A+ \maroonC B(n-1) where A\greenE A is the first term and that B\maroonC B is the common difference.
Since the explicit formula is 118(n1)\greenE{-11}\maroonC{-8}(n-1), we know that the first term is 11\greenE{-11} and the common difference is 8\maroonC{-8}.
In conclusion, A=11\greenE A=\greenE{-11} and B=8\maroonC B=\maroonC{-8}.
5) The explicit formula of an arithmetic sequence is h(n)=1+4nh(n)=1+4n.
Complete the missing values in the recursive formula of the sequence.
{h(1)=Ah(n)=h(n1)+B\begin{cases} h(1)=A\\\\ h(n)=h(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}

In the given recursive formula, A\greenE A is the first term and B\maroonC B is the common difference.
Note that the explicit formula is not given in the standard form A+B(n1)\greenE A+\maroonC B(n-1).
Let's find the first two terms:
  • h(1)=1+41=5h(\blueD 1)=1+4\cdot\blueD 1=5
  • h(2)=1+42=9h(\blueD 2)=1+4\cdot\blueD 2=9
Now we can see that the first term is 5\greenE{5} and the common difference is 4\maroonC{4}.
Therefore, A=5\greenE A=\greenE{5} and B=4\maroonC B=\maroonC{4} .
6) The explicit formula of an arithmetic sequence is i(n)=236ni(n)=23-6n.
Complete the missing values in the recursive formula of the sequence.
{i(1)=Ai(n)=i(n1)+B\begin{cases} i(1)=A\\\\ i(n)=i(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}

In the given recursive formula, A\greenE A is the first term and B\maroonC B is the common difference.
Note that the explicit formula is not given in the standard form A+B(n1)\greenE A+\maroonC B(n-1).
Let's find the first two terms:
  • i(1)=2361=17i(\blueD 1)=23-6\cdot\blueD 1=17
  • i(2)=2362=11i(\blueD 2)=23-6\cdot\blueD 2=11
Now we can see that the first term is 17\greenE{17} and the common difference is 6\maroonC{-6}.
Therefore, A=17\greenE A=\greenE{17} and B=6\maroonC B=\maroonC{-6} .

Challenge problem

7*) Select all the formulas that correctly represent the arithmetic sequence 101,114,127,...101, 114, 127,...
Choose all answers that apply:
Choose all answers that apply:

The first term of the sequence is 101\greenE{101}, and the common difference is 13\maroonC{13}.
Therefore, this is the correct recursive formula for the sequence.
{j(1)=101j(n)=j(n1)+13\begin{cases} j(1)=\greenE{101}\\\\ j(n)=j(n-1)\maroonC{+13} \end{cases}
The other recursive formulas have the wrong first term.
This is the standard form explicit formula for the sequence.
j(n)=101+13(n1)j(n)=\greenE{101}\maroonC{+13}(n-1)
We can simplify it to find another correct formula.
j(n)=101+13(n1)=101+13n13=88+13n\begin{aligned}j(n)&=101+13(n-1)\\\\ &=101+13n-13\\\\ &=88+13n\end{aligned}
The other explicit formulas are not equivalent to the the above formula, and therefore are incorrect.
In conclusion, these are the correct formulas:
  • {j(1)=101j(n)=j(n1)+13\begin{cases} j(1)=101\\\\ j(n)=j(n-1)+13 \end{cases}
  • j(n)=101+13(n1)j(n)=101+13(n-1)
  • j(n)=88+13nj(n)=88+13n
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