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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, comma, 5, comma, 7, comma, point, point, point
a, left parenthesis, n, right parenthesis, equals, 3, plus, 2, left parenthesis, n, minus, 1, right parenthesis
In the formula, n is any term number and a, left parenthesis, n, right parenthesis is the n, start superscript, start text, t, h, end text, end superscript 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, equals, 5 into the explicit formula.
\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, comma, 5, comma, 7, comma, point, point, point

1) 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

Writing explicit formulas

Consider the arithmetic sequence 5, comma, 8, comma, 11, comma, point, point, point The first term of the sequence is start color #0d923f, 5, end color #0d923f and the common difference is start color #ed5fa6, 3, end color #ed5fa6.
We can get any term in the sequence by taking the first term start color #0d923f, 5, end color #0d923f and adding the common difference start color #ed5fa6, 3, end color #ed5fa6 to it repeatedly. Check out, for example, the following calculations of the first few terms.
nCalculation for the n, start superscript, start text, t, h, end text, end superscript term
1start color #0d923f, 5, end color #0d923fequals, start color #0d923f, 5, end color #0d923f, plus, 0, dot, start color #ed5fa6, 3, end color #ed5fa6, equals, 5
2start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 3, end color #ed5fa6equals, start color #0d923f, 5, end color #0d923f, plus, 1, dot, start color #ed5fa6, 3, end color #ed5fa6, equals, 8
3start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 3, plus, 3, end color #ed5fa6equals, start color #0d923f, 5, end color #0d923f, plus, 2, dot, start color #ed5fa6, 3, end color #ed5fa6, equals, 11
4start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 3, plus, 3, plus, 3, end color #ed5fa6equals, start color #0d923f, 5, end color #0d923f, plus, 3, dot, start color #ed5fa6, 3, end color #ed5fa6, equals, 14
5start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 3, plus, 3, plus, 3, plus, 3, end color #ed5fa6equals, start color #0d923f, 5, end color #0d923f, plus, 4, dot, start color #ed5fa6, 3, end color #ed5fa6, equals, 17
The table shows that we can get the n, start superscript, start text, t, h, end text, end superscript term (where n is any term number) by taking the first term start color #0d923f, 5, end color #0d923f and adding the common difference start color #ed5fa6, 3, end color #ed5fa6 repeatedly for n, minus, 1 times. This can be written algebraically as start color #0d923f, 5, end color #0d923f, start color #ed5fa6, plus, 3, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis.
In general, this is the standard explicit formula of an arithmetic sequence whose first term is start color #0d923f, A, end color #0d923f and common difference is start color #ed5fa6, B, end color #ed5fa6:
start color #0d923f, A, end color #0d923f, plus, start color #ed5fa6, B, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis

2) Write an explicit formula for the sequence 2, comma, 9, comma, 16, comma, point, point, point.
d, left parenthesis, n, right parenthesis, equals

3) Write an explicit formula for the sequence 9, comma, 5, comma, 1, comma, point, point, point.
e, left parenthesis, n, right parenthesis, equals

4) The explicit formula of a sequence is f, left parenthesis, n, right parenthesis, equals, minus, 6, plus, 2, left parenthesis, n, minus, 1, right parenthesis.
What is the first term of the sequence?
What is the common difference?

Equivalent explicit formulas

Explicit formulas can come in many forms.
For example, the following are all explicit formulas for the sequence 3, comma, 5, comma, 7, comma, point, point, point
• 3, plus, 2, left parenthesis, n, minus, 1, right parenthesis (this is the standard formula)
• 1, plus, 2, n
• 5, plus, 2, left parenthesis, n, minus, 2, right parenthesis
The formulas may look different, but the important thing is that we can plug an n-value and get the correct n, start superscript, start text, t, h, end text, end superscript 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, comma, 8, comma, 14, comma, point, point, point has a first term of start color #0d923f, 2, end color #0d923f and a common difference of start color #ed5fa6, 6, end color #ed5fa6.
The explicit formula start color #0d923f, 2, end color #0d923f, start color #ed5fa6, plus, 6, end color #ed5fa6, left parenthesis, n, minus, 1, right parenthesis describes this sequence, but the explicit formula start color #0d923f, 2, end color #0d923f, start color #ed5fa6, plus, 6, end color #ed5fa6, n describes a different sequence.
In order to bring the formula 2, plus, 6, left parenthesis, n, minus, 1, right parenthesis to an equivalent formula of the form A, plus, B, n, we can expand the parentheses and simplify:
\begin{aligned}&\phantom{=}2+6(n-1)\\\\ &=2+6n-6\\\\ &=-4+6n\end{aligned}
Some people might prefer the formula minus, 4, plus, 6, n over the equivalent formula 2, plus, 6, left parenthesis, n, minus, 1, right parenthesis, because it's shorter. The nice thing about the longer formula is that it gives us the first term.

5) Find all correct explicit formulas of the sequence 12, comma, 7, comma, 2, comma, point, point, point

Challenge problems

6*) Find the 124, start superscript, start text, t, h, end text, end superscript term of the arithmetic sequence 199, comma, 196, comma, 193, comma, point, point, point

7*) The first term of an arithmetic sequence is 5 and the tenth term is 59.
What is the common difference?

Want to join the conversation?

• what dose it mean to create an explicit formula for a geometric
• An explicit formula directly calculates the term in the sequence that you want.
A recursive formula calculates each term based upon the value of the prior term. So, it usually takes more steps.
• What's the difference between this formula and a(n) = a(1) + (n - 1)d? Is one better or something?
• The reason we use a(n)= a+b( n-1 ), is because it is more logical in algebra
• Can you add a section on Simplifying Geometric and arithmetic equations? I have a test on this tomorrow, and there isn't a section to help me study... Wish me luck I guess :~:
• To find the common difference between two terms, is taking the difference and dividing by the number of terms a viable workaround?

For example, the first term is 5 and the tenth term is 59. So take the difference, 59-5=54, then divide by the number of terms between them, which is 5. And the average difference would be 6. Does this always work?
• Your shortcut is derived from the explicit formula for the arithmetic sequence like 5 + 2(n – 1) = a(n). Plug your numbers into the formula where x is the slope and you'll get the same result:
5 + x(10 – 1) = 59
5 + 9x = 59
9x = 54
x = 6
To find the slope, you take the difference between the 10th and the 1st term and divide it by the "# of additions" or by the difference between the 10th term and the 1st term.
So your shortcut works always, because it's the same thing as the explicit formula (5 + 6(10 – 1) = 59). Basically, it says to get the 10th number in the sequence, you have to start from the base of 5 then add 6 (the slope) to it not once, but in this case 9 times ("# of additions").

P.S. You have a typo in "divide by the number of terms between them, which is 5". It should be 9, not 5. Also the "average difference" should be called "common difference" or "slope".
• The first term of an arithmetic sequence is
5 and the tenth term is 59
How do we make sure the common difference is positive 6 and not negative 6?
• the common difference is constant so we know the sequence will either get bigger or smaller, the tenth term is bigger than the first term so we know it will be positive
• what is the recursive formula for airthmetic formula
• do we need to know arithmetic sequences for the SAT?
• It'll probably help. Most of this kind of stuff is less for real world applications but for creating foundations of understanding for later skills that might require a foundation in understanding and representing sequences correctly. You never know what's ahead.
• Determine the next 2 terms of this sequence
2,5,10,17.. then write the explicit form
this isn't an arithmetic ("linear") sequence because the differences between the numbers are different (5-2=3, 10-5=5, 17-10=7)
however, you might notice that the differences of the differences between the numbers are equal (5-3=2, 7-5=2). that means the sequence is quadratic/power of 2.
since the sequence is quadratic, you only need 3 terms.
let x=the position of the term in the sequence
let y=the value of the term
the 1st term is 2, so x=1 and y=2
the 2nd term is 5, so x=2 and y=5
the 3rd term is 10, so x=3 and y=10
the function is y=ax^2+bx+c, so plug in each point to solve for a, b, and c.
(1,2): 2=a(1^2)+b(1)+c
(2,5): 5=a(2^2)+b(2)+c
(3,10): 10=a(3^2)+b(3)+c

simplify: 2=a+b+c
5=4a+2b+c
10=9a+3b+c

solve this using any method, but i'll use elimination:
10=9a+3b+c
-(5=4a+2b+c)
5=5a+b (equation 3 - equation 2)

5=4a+2b+c
-(2=a+b+c)
3=3a+b (equation 2 - equation 1)

then subtract the 2 equations just produced:
5=5a+b
-(3=3a+b)
2=2a
that means a=1.
substitute a=1 into 3=3a+b: 3=3+b, b=0.
substitute a and b into 2=a+b+c: 2=1+0+c, c=1

so the equation becomes y=1x^2+0x+1, or y=x^2+1
btw you can check (4,17) to make sure it's right
• It seems to me that 'explicit formula' is just another term for iterative formulas, because both use the same form. Is this true? And is there another term for formulas using the m_ + _Bn form as opposed to the A_ + _B(n-1) form or are they both referred to as explicit formulas?
• m + Bn and A + B(n-1) are both equivalent explicit formulas for arithmetic sequences. A + B(n-1) is the standard form because it gives us two useful pieces of information without needing to manipulate the formula (the starting term A, and the common difference B).

An explicit formula isn't another name for an iterative formula. Even though they both find the same thing, they each work differently--they're NOT the same form.

In the iterative formula, "a(n-1)" means "the value of the (n-1)th term in the sequence", this is not "a times (n-1)."

In the explicit formula "d(n-1)" means "the common difference times (n-1), where n is the integer ID of term's location in the sequence."

Thankfully, you can convert an iterative formula to an explicit formula for arithmetic sequences. Converting is usually less work.

Take the iterative formula:
a(1) = A
a(n) = a(n-1) + B (here a(n-1) is this function for the previous term, not multiplication)

Turn it into an explicit formula by taking the initial term's value and adding it to B times the integer (n-1):
a(n) = A + B(n-1) (here B(n-1) is multiplication, not a function)