Explicit formulas for arithmetic sequences

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

How explicit formulas work

Here is an explicit formula of the sequence 3,5,7,...3, 5, 7,...
a(n)=3+2(n1)a(n)=3+2(n-1)
In the formula, nn is any term number and a(n)a(n) is the nthn^\text{th} term.
This formula allows us to simply plug in the number of the term we are interested in, and we will 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}
Cool! This is in fact the fifth term of 3,5,7,...3, 5, 7,...

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Writing explicit formulas

Consider the arithmetic sequence 5,8,11,...5,8,11,... The first term of the sequence is 5\greenE5 and the common difference is 3\maroonC3.
We can get any term in the sequence by taking the first term 5\greenE5 and adding the common difference 3\maroonC3 to it repeatedly. Check out, for example, the following calculations of the first few terms.
nnCalculation for the nthn^\text{th} term
115\greenE{5}=5+03=5=\greenE{5}+0\cdot\maroonC{3}=5
225+3\greenE{5}\maroonC{+3}=5+13=8=\greenE{5}+1\cdot\maroonC{3}=8
335+3+3\greenE{5}\maroonC{+3+3}=5+23=11=\greenE{5}+2\cdot\maroonC{3}=11
445+3+3+3\greenE{5}\maroonC{+3+3+3}=5+33=14=\greenE{5}+3\cdot\maroonC{3}=14
555+3+3+3+3\greenE{5}\maroonC{+3+3+3+3}=5+43=17=\greenE{5}+4\cdot\maroonC{3}=17
The table shows that we can get the nthn^\text{th} term (where nn is any term number) by taking the first term 5\greenE{5} and adding the common difference 3\maroonC{3} repeatedly for n1n\!-\!\!1 times. This can be written algebraically as 5+3(n1)\greenE{5}\maroonC{+3}(n-1).
In general, this is the standard explicit formula of an arithmetic sequence whose first term is A\greenE A and common difference is B\maroonC B:
A+B(n1)\greenE A+\maroonC B(n-1)

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Equivalent explicit formulas

Explicit formulas can come in many forms.
For example, the following are all explicit formulas for the sequence 3,5,7,...3,5,7,...
  • 3+2(n1)3+2(n-1) (this is the standard formula)
  • 1+2n1+2n
  • 5+2(n2)5+2(n-2)
The formulas may look different, but the important thing is that we can plug an nn-value and get the correct nthn^\text{th} term (try for yourselves that the other formulas are correct!).
Different explicit formulas that describe the same sequence are called equivalent formulas.

A common misconception

An arithmetic sequence may have different equivalent formulas, but it's important to remember that only the standard form gives us the first term and the common difference.
For example, the sequence 2,8,14,...2, 8, 14,... has a first term of 2\greenE 2 and a common difference of 6\maroonC 6.
The explicit formula 2+6(n1)\greenE 2\maroonC{+6}(n-1) describes this sequence, but the explicit formula 2+6n\greenE 2\maroonC{+6}n describes a different sequence.
In order to bring the formula 2+6(n1)2+6(n-1) to an equivalent formula of the form A+BnA+Bn, we can expand the parentheses and simplify:
=2+6(n1)=2+6n6=4+6n\begin{aligned}&\phantom{=}2+6(n-1)\\\\ &=2+6n-6\\\\ &=-4+6n\end{aligned}
Some people might prefer the formula 4+6n-4+6n over the equivalent formula 2+6(n1)2+6(n-1), because it's shorter. The nice thing about the longer formula is that it gives us the first term.

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