Arithmetic sequences - Sequences and Series

What is a sequence?

A sequence is a list of things placed in a certain order. This lesson will teach you how to solve questions relating to sequences.

What is an arithmetic sequence

When you come across an arithmetic sequence, the difference between one term and the next one is constant. So for example, when we have a sequence of 1,2,3,4,51, 2 , 3, 4, 5, this is arithmetic since each number is simply +1 to the previous number. There’s actually a special name for the +11 in this situation…

What is common difference

What is the common difference definition? When you figure out the constant difference between terms in an arithmetic sequence, you’ve found the common difference! In the previous example, 11 is the common difference between the terms.

How to find the common difference

When you’ve got an arithmetic sequence, how do you find the common difference? Firstly, your sequence should be in order. If you know that the sequence is arithmetic, then simply find the difference between the 2nd2^{nd} and 1st1^{st} term and you’ll find the common difference. If you wanted to check your answer, feel free to find the difference between any consecutive terms and you should find that you’ll get the same answer for your common difference.

On the other hand, if you’re not sure if a sequence is arithmetic, but you find that there’s a constant difference between all the terms, you’ve proven that the sequence is arithmetic.

How to find the nthn^{th} term

When you are trying to find the nthn^{th} term of an arithmetic sequence, use the following arithmetic sequence formula:

tn=t1+(n1)dt_{n} = t_{1} + (n - 1)d

tnt_{n} is the nthn^{th} term, and t1t_{1} is the first term. dd is the common difference in the arithmetic sequence. So if we use the example from above where we have a sequence of 1,2,3,4,51, 2 , 3, 4, 5, if we had to find what would be the 6th6^{th} term to come after the number 55 in our sequence, we’d get:

t6=1+(61)1t_{6} = 1 + (6 - 1)1

t6=6t_{6} = 6

This makes sense since we know the next number to come should be 66. Use the above formula and it will help you in finding the nth term. Let’s look at more examples on arithmetic sequences.

Example problems

Question 1:

Arithmetic sequence formula

Consider the arithmetic sequence: 5, 9, 13, 17, …

a) Identify the common difference.


Common difference is the difference between successive terms. We can pick any pair of successive terms to calculate the common difference. In this question, we have four terms:

t1=5,t2=9,t3=13,t4=17t_{1} = 5, t_{2} = 9, t_{3} = 13, t_{4} = 17

We can pick any two successive terms from here. So:

d=t2t1=95=4d = t_{2} - t_{1} = 9 - 5 = 4

d=t3t2=139=4d = t_{3} - t_{2} = 13 - 9 = 4

d=t4t3=1713=4d = t_{4} - t_{3} = 17 - 13 = 4

Looking for the common difference (d) in arithmetic sequence
Look for common difference

We now know the common difference of this arithmetic sequence is 44

b) Determine the seventh term of the sequence.


We can use this equation to find the answer:

tn=t1+(n1)dt_{n} = t_{1} + (n - 1)d

tn=ntht_{n} = n^{th} term

t1=1stt_{1} = 1^{st} term

dd: common difference

We are looking for the seventh term. So:

tn=t1+(n1)dt_{n} = t_{1} + (n - 1)d

t7=5+(71)4t_{7} = 5 + (7 - 1)4

t7=5+24t_{7} = 5 + 24

t7=29t_{7} = 29

We can verify the answer by finding the terms up to the seventh term one by one. To do that, we just need to add the common difference to the last term:

t4=17,t5=17+4=21,t6=21+4=25,t7=25+4=29t_{4} = 17, t_{5} = 17 + 4 = 21, t_{6} = 21 + 4 = 25, t_{7} = 25 + 4 = 29

c) Which term in the sequence has a value of 8585?


We can use this equation to find the answer:

tn=t1+(n1)dt_{n} = t_{1} + (n - 1)d

We know the tn,t1t_{n}, t_{1}, and dd. So we put the value into the equation

85=5+(n1)485 = 5 + (n - 1)4

Now we solve for nn

85=5+(n1)485 = 5 + (n - 1)4

80=(n1)480 = (n - 1)4

804=(n1)44\frac{80}{4} = \frac{(n - 1)4}{4}

20=n120 = n - 1

n=21n = 21

Question 2:

Determine t1,d,tnt_{1}, d, t_{n} for the sequences in which two terms are given

t4=14,t10=32t_{4} = 14, t_{10} = 32


We can use this equation as a starting point to find the answer:

tn=t1+(n1)dt_{n} = t_{1} + (n - 1)d

We put the two terms we know to the equation:

t4=t1+(41)d=14=t1+3dt_{4} = t_{1} + (4 - 1)d = 14 = t_{1} + 3d

t10=t1+(101)d=32=t1+9dt_{10} = t_{1} + (10 - 1)d = 32 = t_{1} + 9d

Now we have two equations:

14=t1+3d14 = t_{1} + 3d

32=t1+9d32 = t_{1} + 9d

We can use elimination to solve for dd and t1t_{1}

14=t1+3d14 = t_{1} + 3d

()32=t1+9d(-) 32 = t_{1} + 9d

18=6d-18 = -6d

d=3d = 3

We put d=3d = 3 into either one of the equations

14=t1+3(3)14 = t_{1} + 3(3)

14=t1+914 = t_{1} + 9

t1=5t_{1} = 5

We can find tnt_{n} using the information we have

tn=t1+(n1)dt_{n} = t_{1} + (n - 1)d

tn=5+(n1)3t_{n} = 5 + (n - 1)3

tn=5+3n3t_{n} = 5 + 3n - 3

tn=3n+2t_{n} = 3n + 2

Want to double check your answers? Click here for a handy arithmetic sequence calculator.

Looking to move forward? Learn how to use pascal’s triangle or what the binomial theorem is.

Or expand on sequences and learn about monotonic and bounded sequences.

Arithmetic sequences

An arithmetic sequence (arithmetic progression) is a number sequence with a common difference between successive terms. By using the arithmetic sequence formula, we can easily find the value of a term and the common difference in the sequence.


• arithmetic sequence: a sequence with a common difference between successive terms
• The nth term, tn{t_n} ,of an arithmetic sequence:
tn=t1+(n1)d{t_n} = {t_1} + \left( {n - 1} \right)d
where, tn{t_n}: nth term
t1{t_1}: first term
dd : common difference
  • 1.
    Arithmetic sequence formula
    Consider the arithmetic sequence: 5, 9, 13, 17, … .
  • 2.
    Determine t1,d,tnt_1,d,t_n for the sequences in which two terms are given
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Arithmetic sequences

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